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ArbitrLikvidat [17]
3 years ago
6

You drop a stone down off a bridge. You are able to count to 4.0 seconds when it finally hits the water. How high is the bridge?

​
Physics
1 answer:
mart [117]3 years ago
7 0

Answer:

The height of the bridge is 78.4 m.

Explanation:

Given;

time of the stone motion off the bridge, t = 4.0 s

acceleration due to gravity, g = 9.8 m/s²

The height of the bridge is given by;

h = ut + ¹/₂gt²

where;

u is the initial velocity of the stone, u = 0

h = ¹/₂gt²

h = ¹/₂(9.8)(4)²

h = 78.4 m

Therefore, the height of the bridge is 78.4 m.

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Answer:

0.0031792338 rad/s

Explanation:

\theta = Angle of elevation

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Using trigonometry

tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta

Differentiating with respect to t we get

\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft

Now y = 2500 ft

cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974

\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s

The angle is changing at 0.0031792338 rad/s

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Answer:

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We will use conversation of energy.

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Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

K_B_1 + U_B_1 = K_B_2 + U_B_2

\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B

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Explanation:

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Compare mass and weight
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