Question

Two students are on a balcony 21.9 m above the street. One student throws a ball vertically downward at 13.6 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down. What is the magnitude of the velocity of the first ball as it strikes the ground?

Answer 1

Answer: 24.77 m/s

Explanation: Knowing the following:

Initial Velocity (Vo) = 13.6 m/s

Gravity (g) = 9.8 $m/2^{2}$

total distance = 21.9m

To know the final velocity (Vf) on which the ball drops the ground, we can just sum the initial velocity (Vo) plus the velocity gained through the gravity:

Vf = Vo + g*t

However, we dont know the time the takes to touch the ground. But we can get it using the following.

D = Vo * t0 + ($\frac{1}{2}$* g * $t^{2}$)

This is a cuadratic equation that can be solved in several ways. But first, we need to acomadate this equation

21.9 = 13.6 * t + 4.9* $t^{2}$

4.9 $t^{2}$ + 13.6 * t -21.9 = 0

On quadratic equation, we can know the value for time (t) by running the following equation:

±$t = \frac{-b ± \sqrt{b^{2} - 4ac}}{2a}$

Where:

a = 4.9

b = 13.6

c= -21.9

Running this equation we get the following values:

t = 1.14 ;−3.91

We take the positive time, because time runs only one way. With this time we do the equation:

Vf = Vo + g*t

Which give us:

Vf = 13.6 + 9.8 * 1.14

Vf= 24.77 m/s