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FinnZ [79.3K]
3 years ago
5

A ball is thrown horizontally at a speed of 16 m/s from the top of a cliff. If the ball hits the ground 6.0 s later, approximate

ly how high is the cliff? 360m 240m 180m 150m
Physics
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

Y = 176.4 m

Explanation:

For the height of cliff we will analyze the vertical motion. We will apply the 2nd equation of motion:

Y = V₀y*t + (0.5)gt²

where,

Y = Height = ?

V₀y = Initial Vertical Velocity = 0 m/s (since, ball is thrown horizontally)

t = time = 6 s

g = 9.8 m/s²

Therefore,

Y = (0 m/s)(6 s) + (0.5)(9.8 m/s²)(6 s)²

<u>Y = 176.4 m</u>

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Missing details. Complete text is:"The following reaction has an activation energy of 262 kJ/mol:
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\ln( \frac{K_2}{K_1} ) =  \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}  )
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