A large building has an inclined roof. The length of the roof is 47.5 m and the

Question

A large building has an inclined roof. The length of the roof is 47.5 m and the angle of the roof is 24.0° below horizontal. A worker on the roof lets go of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 3.99 m/s2. After leaving the edge of the roof, it falls a vertical distance of 43.5 m before hitting the ground.

(a) How much time does it take the hammer to fall from the edge of the roof to the ground?

(b) How far horizontally does the hammer travel from the edge of the roof until it hits the ground?

Explanation / Answer

according to motion equations

V^2 = U^2 + 2*a*S

where U is intial velocity and V is final velocity and 'a' is acceleration and S is length of the roof

=> V^2 = 0+ 2*3.99*47.5 = 379.05

=> V = 19.46 m/s at the end of the roof .

Time taken by hammer to travel the roof length is t1.

=> v = u + at

=> t1 = v/a = 19.46/3.99 = 4.87 sec

At the end of the roof

velocity of hammer in vertical direction Vv = Vsin(24 degree) = 19.46* 0.406 =7.9 m/s

=> time taken by hammer in vertical direction t2 = 43.5/7.9 = 5.50 sec

SO (a) total time T = t1+t2 = 5.50+4.87 = 10.37 sec

(b) velocity of hammer in hprizontal direction Vh = Vcos(24 degree) = 19.46*0.913 = 17.76 m/s

SO horizontal distance travelled by hammer = 17.76* t2 = 17.76 *5.50 = 97.68 m   

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