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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.

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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
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If $g$ denotes the gravity, $u$ the initial speed, $d$ the travelled horizontal distance, $x\in \left[0,\frac\pi2\right]$ the angle to the ground of the initial velocity, and $t$ the flight time, then you have $u \cos(x) t= d$ and $u\sin(x) t-\frac12gt^2=0$. Solve for $x$ to get $\sin(2x)=\frac{gd}{u^2}$. There are two possible values for $x$, whence also two values for $t=\frac{d}{u\cos(x)}$.

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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.

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  • $\begingroup$ 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

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