Answer :
Answer:
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!