You throw a ball upward from ground level with initial upward speed v0. What is the max height of the trajectory?

How long does it take the ball to impact the ground?

What is the speed of the ball when it hits the ground?

Answer with mathematical expression and state all assumptions.

The max height of the ball is y = - 1/2 (v0²/g).

It takes the ball t = - 2 · v0/g to hit the ground.

The speed of the ball when it hits the ground is v = - v0.

Explanation:

The height and velocity of the ball is given by the following equations:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the ball at time t

y0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t

When the ball is at max height, the velocity is 0. So, let's find the time at which the velocity of the ball is 0.

v = v0 + g · t

0 = v0 + g · t

t = - v0/g

Now, replacing t = - v0/g in the equation of height, we will obtain the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located on the ground)

y = v0 · t + 1/2 · g · t²

Replacing t:

y = v0 · (-v0/g) + 1/2 · g · (-v0/g) ²

y = - (v0²/g) + 1/2 · (v0²/g)

y = - 1/2 (v0²/g)

The max height of the ball is y = - 1/2 (v0²/g). Remember that g is negative.

Since the acceleration of the ball is always the same, the time it takes the ball to impact the ground will be twice the time it takes to reach its max height, t = - 2 v0/g.

However, let's calculate that time knowing that at that time the height is 0:

y = y0 + v0 · t + 1/2 · g · t²

0 = v0 · t + 1/2 · g · t²

0 = t · (v0 + 1/2 · g · t)

0 = v0 + 1/2 · g · t

-2 · v0/g = t

It takes the ball t = - 2 · v0/g to hit the ground.

Let's use the equation of velocity at final time (t = - 2 · v0/g):

v = v0 + g · t

v = v0 + g · (-2 · v0/g)

v = v0 - 2· v0

v = - v0

The speed of the ball when it hits the ground is v = - v0.