Speed and height reached

Question

A ball player catches a ball 3.4 s after throwing it vertically upward. With what speed did he throw it, and what height did it reach?

Answer 1

If he caught the ball 3.4 seconds after throwing it, it would have been going up for 1.7 s and coming down for 1.7 s. At the peak of travel, instantaneously it's velocity is zero because it goes from having a positive velocity to having a negative velocity and it cannot do so without passing through zero.

The equation of motion we can use is :

v_f = v_i + at where "v_f" is final velocity (which is zero), v_i is initial velocity, "a" is acceleration (which is -9.8 m/s² in this case), and "t" is time. Plugging in the numbers we get:

0 = v_i + (-9.8)(1.7)

0 = v_i - 16.66

v_i = 16.66 m/s

To find out how high it went, we can use the equation of motion:

y = v_it + 1/2at²

y = (16.66)(1.7) + 1/2(-9.8)(1.7²)

y = 28.322 - 14.161

y = 14.161 m

I'll let you round to the appropriate number of sig figs.