A ball player catches a ball 3.3 seconds after throwing it vertically upward. With what speed did he throw it,

Question

and what height did it reach?

Answer 1

the only given are the time = 3.3 sec and g = 9.8 m/s^2

at highest point final velocity is zero.
so initial velocity = vf - gt
vi = 0 - gt
since the motion of the ball is upward and acceleration(g) is downward, so g is negative
vi = 0 - (-9.8m/s^2)t
vi =(9.8m/s^2)(3.3sec/2)
vi = 16.17 m/sec

d = (vf^2- vi^2)/ 2g
d = (0^2 - 16.17^2) / 2*-9.8
d = (0 -261.47)/(-19.6)
d = 13.34 m

hope this helps you friend

Answer 2

The ball takes as much time going up as coming down. Hence t1 = t/2 = 3.3/2 = 1.65 sec.

Now the distance travelled from the top to the ground is:

y = -1/2 g*t1^2 = 13.34 m --> Note the ball's speed at the top of the path is zero.

The speed of the ball at the ground is:

v = gt = 9.8 *3.3/2 = 16.7 m/s

Answer 3

If you want to be really precise in your answer, the maximum height reached by the ball depends on the player's height. Otherwise, the answer given previously is correct.