Question

If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hits the ground at the end of its flight?

Answer 1

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) =  (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 =  (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 =   (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s