Find the initial direction and time of flight of a basketball, given initial speed and distance
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A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?
I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.
algebra-precalculus trigonometry physics
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
$endgroup$
add a comment |
$begingroup$
A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?
I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.
algebra-precalculus trigonometry physics
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
$endgroup$
$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13
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I didn't understand that, what's the latter interperatation
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– Zayn Malek
Oct 18 '15 at 0:17
$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20
add a comment |
$begingroup$
A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?
I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.
algebra-precalculus trigonometry physics
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
$endgroup$
A player passes a basketball to another player who catches it at the same level from which it was thrown. The initial speed of the ball is 7.1m/s, and it travels a distance of 4.6m. What were (a) the initial direction of the ball and (b) the time of flight?
I can't figure a way using only kinematic equations and soh cah toa, am I missing something? I tried using trigonometric identities but got stumped late into the algebra.
algebra-precalculus trigonometry physics
algebra-precalculus trigonometry physics
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
edited Oct 18 '15 at 5:06
user147263
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
asked Oct 17 '15 at 23:47
Zayn MalekZayn Malek
63
63
$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13
$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17
$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20
add a comment |
$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13
$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17
$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20
$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13
$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13
$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17
$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17
$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20
$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
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add a comment |
$begingroup$
Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is
$$ frac{distance }{v cos alpha} $$
So $alpha$ should also be known.
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
$endgroup$
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
$endgroup$
add a comment |
$begingroup$
If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
$endgroup$
add a comment |
$begingroup$
If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
$endgroup$
If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $xin left[0,fracpi2right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u cos(x) t= d$ and $usin(x) t-frac12gt^2=0$. Solve for $x$ to get $sin(2x)=frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=frac{d}{ucos(x)}$.
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
answered Oct 18 '15 at 0:20
BatominovskiBatominovski
33.1k33293
33.1k33293
add a comment |
add a comment |
$begingroup$
Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is
$$ frac{distance }{v cos alpha} $$
So $alpha$ should also be known.
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
$endgroup$
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
add a comment |
$begingroup$
Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is
$$ frac{distance }{v cos alpha} $$
So $alpha$ should also be known.
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
$endgroup$
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
add a comment |
$begingroup$
Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is
$$ frac{distance }{v cos alpha} $$
So $alpha$ should also be known.
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
$endgroup$
Is the ball is thrown along a parabolic path at an angle $ alpha $ with speed $v$, the time is
$$ frac{distance }{v cos alpha} $$
So $alpha$ should also be known.
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
answered Oct 18 '15 at 0:03
NarasimhamNarasimham
20.9k62158
20.9k62158
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
add a comment |
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
The theta wasn't given.
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:05
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
If the ball is thrown up at different angles, the times taken will be different. Almost a vertical throw takes a longer time,right?
$endgroup$
– Narasimham
Oct 18 '15 at 0:09
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Ooh I get that. How did you derive that equations
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:13
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
$begingroup$
Read up about projectile motion, how to find the trajectory, time velocity etc. It is interesting that horizontal velocity is not influenced by gravity. Accept also to learn such interesting things.
$endgroup$
– Narasimham
Oct 18 '15 at 0:32
add a comment |
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$begingroup$
Is 4.6 meters the distance travelled in air, or the horizontal displacement? The latter interpretation makes the problem much simpler.
$endgroup$
– Semiclassical
Oct 18 '15 at 0:13
$begingroup$
I didn't understand that, what's the latter interperatation
$endgroup$
– Zayn Malek
Oct 18 '15 at 0:17
$begingroup$
It's what @C.I.J. assumes in his solution
$endgroup$
– Semiclassical
Oct 18 '15 at 0:20