020 (part 1 of 3) 10.0 points A student stands at the edge of a cliff and throws

Question

020 (part 1 of 3) 10.0 points A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 15.3 m/s. The cliff is 75.8 m above a flat, horizontal beach. 15.3 m/s How long after being released does the stone strike the beach below the cliff? The acceler ation of gravity is 9.8 m/s Answer in units of s. 021 (part 2 of 3) 10.0 points At impact, what is its speed? Answer in units of m/s. 022 (part 3 of 3) 10.0 points At what angle below the horizontal does it land? Answer in units of

Explanation / Answer

Here , for the initial speed ,

u = 15.3 m/s

height , h = 75.8 m

a) let the time taken is t

Using second equation of motion

d = uvertical * t + 0.50 at^2

75.8 = 0.50 * 9.8 * t^2

t = 3.933 s

the time of flight is 3.93 s

b)

at the impact speed is v

Using third equation of motion

v^2 - u^2 = 2 * a * d

v^2 - 15.3^2 = 2 * 9.8 * 75.8

solving for v

v = 41.5 m/s

the speed is 41.5 m/s

3)

let the angle is theta

41.5 * cos(theta) = 15.3

cos(theta) = 0.37

theta = 68.4 degree

the angle below horizontal the is 68.4 degree

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