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1 answer:
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Given:
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
v = 50.0 m/s, the launch velocity
θ = 36.9°, the launch angle above the horizontal
Assume g = 9.8 m/s² and ignore air resistance.
The vertical component of the launch velocity is
Vy = (50 m/s)*sin(50°) = 30.02 m/s
The time, t, to reach maximum height is given by
(30.02 m/s) - (9.8 m/s²)*(t s) = 0
t = 3.0634 s
The time fo flight is 2*t = 6.1268 s
The horizontal velocity is
u = (50 m/s)cos(36.9°) = 39.9842 m/s
The horizontal distance traveled at time t is given in the table below.
Answer:
t, s x, m
------ --------
0 0
1 39.98
2 79.79
3 112.68
4 159.58
5 199.47
6 239.37
Answer:
The car stops after 32.58 m.
Explanation:
t = Time taken for the car to stop
u = Initial velocity = 20 m/s
v = Final velocity = 0
s = Displacement
a = Acceleration = -6 m/s²
Time taken by the car to stop
Total Time taken by the car to stop is 0.5+3.33 = 3.83 s
The car stops after 32.58 m.
Distance between car and obstacle is 50-32.58 = 17.42 m