Question

A ball is thrown upward with an initial vertical velocity of v0 to a maximum height of hmax and falls back toward earth. on the way down, it is caught at the same height at which it was thrown upward. b b b b b b b b b bbb b b b b b b b b v0 hmax if the total time the ball remains in the air is t, find its speed when caught. the acceleration of gravity is g. neglect air resistance.

Answer 1

Answer: $v_f=g\frac{t}{2}$

The ball was thrown at the speed of $v_o$.

Maximum height achieved is $h_{max}$

Time of flight is t.

Now, the time the ball takes to achieve maximum height = the time taken by ball to fall back = $\frac{t}{2}$

let us just consider the second half of the flight. At $h_{max}$, the velocity would be zero. let us consider as the initial velocity for the second half of the flight i.e. $v_i=0$

Using the equation of motion:

$v_f=v_i+at$

where,  $v_f$ is the final velocity, a is the acceleration, t is the time taken.

Because the ball would fall under gravity, hence a=g and time of flight would be t/2

$\Rightarrow v_f=0+g\frac {t}{2}$