While driving north at 21 m/s during a rainstorm you notice that the rain makes an angle of 36° with the vertical. while drivin
g back home moments later at the same speed but in the opposite direction, you see that the rain is falling straight down. from these observations, determine the speed and the angle of the raindrops relative to the ground?
1 answer:
Let the velocity of the vehicle be
V₁ = (21, 0)
Let the velocity of a raindrop be
V₂ = (a, b)
When returning, the raindrop is vertical.
Therefore a = 21 m/s
When driving north, a raindrop makes an angle of 36° with the vertical, therefore
tan(36°) = a/b = 21/b
b = 21/tan(36°) = 28.9 m/s
The speed of a raindrop is
√(a² + b²) = √(441 + 835.44) = 35.73 m/s
The angle is 90-36 = 54° with the ground.
Answer: 35.7 m/s at 54° relative to the ground.
V₁ = (21, 0)
Let the velocity of a raindrop be
V₂ = (a, b)
When returning, the raindrop is vertical.
Therefore a = 21 m/s
When driving north, a raindrop makes an angle of 36° with the vertical, therefore
tan(36°) = a/b = 21/b
b = 21/tan(36°) = 28.9 m/s
The speed of a raindrop is
√(a² + b²) = √(441 + 835.44) = 35.73 m/s
The angle is 90-36 = 54° with the ground.
Answer: 35.7 m/s at 54° relative to the ground.
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Refer to the diagram shown below.
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).
When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.
The time, t, for the bolt to fall a known distance obeys the equation
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.
Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s
The total time, T, to fall 94 m is given by
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s
The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s
The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s
The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s
Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).