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yKpoI14uk [10]
3 years ago
12

A marble is thrown horizontally with a speed of 15.0 m/s from the top of a building. When it strikes the ground, the marble has

a velocity that makes an angle of 65.0° with the horizontal. From what height above the ground the marble was thrown. How far from the base of the building, did the marble land?
Physics
2 answers:
alina1380 [7]3 years ago
8 0
The answer for this question is 52.79
jolli1 [7]3 years ago
3 0
The answer is 52.79 m.

I used this formula to get the formula for Vy:
v^2=vi^2+2(a)(x)
And got
Vy=square root (19.6 h)

Then I used that and put it in this formula:

tan(65) =Vy/Vx

tan(65) = square root (19.6 h)/15.0

Then I rearranged it to:

h=[(15.0)(tan65)]/19.6

h=52.79 m
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Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

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t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

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