Acannon 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.

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

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!

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