Jimmy is standing 300 feet away from a rocket that is being shot up into the air away from him at an angle of elevation of 70 de
grees. Jimmy’s line of sight to the rocket has an angle of elevation of 35 degrees. If he is standing 300 feet from the rocket what is the vertical height of the rocket?
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Answer:
The total time for the race is 11.6 seconds
Explanation:
The parameters given are;
Total distance ran by Willie = 100.0 m
Initial acceleration = 2.00m/s²
Top speed reached with initial acceleration = 12.0 m/s
Point where Willie start to fade and decelerate = 16.0 m from the finish line
Speed with which Willie crosses the finish line = 8.00 m/s
The time and distance covered with the initial acceleration are found using the following equations of motion;
v = u₀ + a·t
v² = u₀² + 2·a·s
Where:
v = Final velocity reached with the initial acceleration = 12.0 m/s
u₀ = Initial velocity at the start of the race = 0 m/s
t = Time during acceleration
a = Initial acceleration = 2.00 m/s²
s = Distance covered during the period of initial acceleration
From, v = u₀ + a·t, we have;
12 = 0 + 2×t
t = 12/2 = 6 seconds
From, v² = u₀² + 2·a·s, we have;
12² = 0² + 2×2×s
144 = 4×s
s = 144/4 =36 meters
Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;
Distance covered at top speed = 100 - 36 - 16 = 48 meters
Time, of running at top speed = Distance/velocity = 48/12 = 4 seconds
The deceleration from top speed to crossing the line is found as follows;
v₁² = u₁² + 2·a₁·s₁
Where:
u₁ = v = 12 m/s
v₁ = The speed with which Willie crosses the line = 8.00 m/s
s₁ = Distance covered during decelerating = 16.0 m
a₁ = Deceleration
From which we have;
8² = 12² + 2 × a × 16
64 = 144 + 32·a
64 - 144 = 32·a
32·a = -80
a = -80/32 = -2.5 m/s²
From, v₁ = u₁ + a₁·t₁
Where:
t₁ = Time of deceleration
We have;
8 = 12 + (-2.5)·t₁
t₁ = (8 - 12)/(-2.5) = 1.6 seconds
The total time = t + + t₁ =6 + 4 + 1.6 = 11.6 seconds.