Question

A ball is thrown straight up at a speed of 2 m/s and is caught at some time later. Assume the ball is thrown and caught at the same height.
(c) What the the ball’s maximum height with respect to it’s initial position?
(d) How long does the ball take to get to it’s maximum height?
(e) What is the ball’s total flight time?
(f) What is the velocity of the ball when it is caught?
(g) Graph the position, velocity, and acceleration of the ball.

Answer 1

(c)Using \[v ^{2}=u^{2}-2gs\] remember \[g=-a\] we know that final velocity \[v=0\] at maximum height, so \[s=\frac{u^{2} }{ 2g}\] take \[g=9.8ms^{-2}\]therefore \[s=0.204m\]
(d)Using \[v=u-gt\] we still have \[v=0\] at maximum height therefore \[t=\frac{u}{g}=0.204s\]
(e)total time of flight is \[2t=2 \times 0.204=0.408s\]
(f)Using \[v=u+gt\] here \[a=g\] and since \[u=0\] at final position we use the second half of the time of flight \[0.204s\] and get \[v=gt=9.8\times0.204=1.9992\approx2ms^{-1}\]
(g)The graphing is easy with the data