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Elis [28]
3 years ago
14

A ball was kicked upward at a speed of 64.2 m/s. How fast was the ball going 1.5 seconds later

Physics
1 answer:
Kobotan [32]3 years ago
3 0

The ball's velocity v at time t is

v=v_0-gt

where g=9.81\,\frac{\mathrm m}{\mathrm s^2} is the acceleration due to gravity. So after 1.5 seconds, the ball's velocity is

v=64.2\,\dfrac{\mathrm m}{\mathrm s}-\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1.5\,\mathrm s)

\implies v=49.5\,\dfrac{\mathrm m}{\mathrm s}

so the ball's speed after 1.5 seconds is also 49.5 m/s.

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Answer:

θ_c = 36.87°

Explanation:

Index of refraction for index medium; n_i = 2

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Formula to find the critical angle is given;

n_i(sin θ_c) = n_r(sin 90)

Where θ_c is critical angle.

Thus;

2 × (sin θ_c) = 1.2 × 1

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(sin θ_c) = 0.6

θ_c = sin^(-1) 0.6

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