Question

A tennis player tosses a tennis ball straight up and then catches it after 1.62 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?

(b) What is the velocity of the ball when it reaches its maximum height?

(c) Find the initial velocity of the ball

(d) Find the maximum height it reaches.

Answer 1

a) acceleration of the ball is g = 9.8 m/s^2. If you choose the upward is positive direction, g = -9.8 m/s^2
b) when it gets maximum height, velocity =0
c) to find out the initial velocity of the ball, you need the formula to find out the distance the ball travels and then set x = x_0 =0 to find the v_0 . It's \[x = x_0 + v_0t + \frac{1}{2}at^2\] in this case a = -g =-9.8 so, \[v_0 t= 4.9t^2 \\v_0 = 4.9t = 4.9* 1.62= 7.9m/s^2\]
d) the maximum height is v =0 , so, apply formula \[v^2=v_0^2 +2as\] again, a = -g = -9.8 , \(v_0^2 = 2*9.8 s\rightarrow s = \dfrac{v_0^2}{19.6}= \dfrac{19.6}{62.41}=3.2m\)
I make the answers in significant figure form, therefore, they are not exactly what the calculator said.

Answer 2

You should check, and recheck. I am new in physics. I've started my physics course 2 weeks ago. hehehe... And this is the first time I study physics.