Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
Answer:
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