A large building has an inclined roof. The length of the roof is 70.0 m and the

Question

A large building has an inclined roof. The length of the roof is 70.0 m and the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 4.14 m/s2. After leaving the edge of the roof, it falls a vertical distance of 44 5 m before hitting the ground (a) How much time, in seconds, does it take the hammer to fall from the edge of the roof to the ground? 2 97 Given the acceleration of the hammer as it sides on the roof, what are the velocity components just as it leaves the roof? How can you use these velocity components to fied the time it takes to fall from the roof to the ground? What constant acceleration relations will you need? s (b) How far horizontally, in meters, does the hammer travel from the edge of the roof until it hits the ground?

Explanation / Answer

(a) Here time taken by the hammer to the edge of the roof = t = [2.s/a] = [(2 * 70) / 4.14] = 5.81 s
So, speed leaving egge of roof = 5.81 × 4.14 = 24.05 m/s
Now, the vertical component of the velocity = 24.05 × sin(25) = 10.16 m/s

Speed when it reaches the ground (from kinetic energy)
= [2.(½ × 10.16² + 44.5 × 9.81)] = sqrt [ 2*(51.61 + 436.55)] = 31.25 m/s
Average speed during fall = (31.25 + 10.16)/2 = 20.70 m/s

Therefore, time taken by the hammer to fall from the edge of the roof to the ground = 44.5 / 20.70 = 2.15 s

(b) Horizontal speed leaving roof = 24.05 × cos(25) = 21.80 m/s

therefore, the horiontal distance travelled by the hammer from edge of roof after 2.15 s = 2.15 × 21.80 = 46.87 m

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