Question

A cannon directed straight upward launches a ball with an initial speed v. The ball reaches a maximum height h in a time t. Then, the same cannon is used to launch a second ball straight upward at a speed 2v. In terms of h and t, what is the maximum height the second ball reaches and how long does it take to reach that height? Ignore any effects of air resistance.

Answer 1

The maximum height of the second ball is 4h and the time it takes the ball to reach that height is √-2h/g

Step-by-step explanation:

Hi there!

The height and velocity of the ball can be calculated using the following equations:

h = h0 + v · t + 1/2 · g · t²

vel = v + g · t

Where:

h = height of the ball at time t.

h0 = initial height.

v = initial velocity.

t = time.

g = acceleration due to gravity.

vel = velocity at time t.

When the ball reaches its maximum height, its velocity is zero:

t₁ = time it takes the first ball to reach its maximum height.

vel = v + g · t₁

0 = v + g · t₁

- g · t₁ = v

At maximum height the initial velocity of the ball is (-g · t₁).

Placing the origin of the frame of reference on the ground so that h0 = 0, the maximum height will be:

h = (-g · t₁) · t₁ + 1/2 · g · t₁²

h = -g · t₁² + 1/2 · g · t₁²

h = -1/2 · g · t₁²

Now, if the initial velocity is 2v, at maximum height:

t₂ = time it takes the second ball to reach its maximum height,

vel = 2v + g · t₂

0 = 2v + g · t₂ (v = -g · t₁)

0 = 2(-g · t₁) + g · t₂

2 · g · t₁ / g = t₂

2 · t₁ = t₂

The maximum height of the ball will be:

h₂ = 2 · v · t₂ + 1/2 · g · t₂² (v = -g · t₁) (t₂ = 2 · t₁)

h₂ = -2 · g · t₁ · (2·t₁) + 1/2 · g · (2 · t₁)²

h₂ = -4 · g · t₁² + 2 · g · t₁²

h₂ = -2 · g · t₁² (h₁ = -1/2 · g · t₁²)

h₂ = 4 h₁

Then:

4 h₁ = -2 · g · t₁²

-2 h₁/g = t₁²

√-2h₁/g = t₁

The maximum height of the second ball is 4h and the time it takes the ball to reach that height is √-2h/g

Have a nice day!