A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge

Question

A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 39.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. Express your answer with the appropriate units. Significant Figures Feedback: Your answer 38.64 ^ was either rounded differently or used a different number of significant figures than required for this part. If you need this result for any later calculation in this item, keep all the digits and round as the final step before submitting your answer. How much time elapses from when the rock is thrown until it hits the street? Express your answer with the appropriate units.

Explanation / Answer

here,

velocity of thrawn, v = 27 m/s
height of building, h = 39 m

Net taken by rock to reach street,
t = time taken to reach max height after thrown + time taken to level at original position + Time taken to reach street,

t = t1 + t2 + t3

From first equation of motion,
v = u + at1

solving for time t1,
t1 = v/g
t1 = 27/9.81
t1 = 2.752 s = t2 ( as same rock levelled original postion)

now from second equation of motion solvign for time t3,

h = 0 + 0.5 * gt2^2

t3 = sqrt(2h/g)
t3 = sqrt(2*39/9.81)
t3 = 2.82 s

Net time taken. t = t1 + t2 + t3
t = 2.752 + 2.752 + 2.82
t = 8.324 s

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