A man stands on the roof of a building of height 16.6 m and throws a rock with a

Question

A man stands on the roof of a building of height 16.6 m and throws a rock with a velocity of magnitude 25.6 m/s at an angle of 32.5 degrees above the horizontal. You can ignore air resistance.

a. Calculate the maximum height above the roof reached by the rock.

b. Calculate the magnitude of the velocity of the rock just before it strikes the ground.

c. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Explanation / Answer

a) initial vertical velocity of rock = 13.75 m/sec time it takes to rise to hmax = 13.75/ g= 1.4 sec hmax= 13.75^2 /(2g) =9.64m above the building b)rock falls from = 9.64+ 16.6 = 26.24 m v^2= u^2 + 2*g*h [ u=0,h= 26.24] =>v= 22.68 m/sec in vertical direction horizontal velocity = 25.6 cos(32.5)=21.59 m/sec so magnitude =31.31 m/sec c)net time when rock travels horizontal distance = time taken to fall back + time taken to rise time taken to fall bac= 22.68/9.8=2.314 sec time taken to rise = 1.4 sec total time = 3.71 sec so horizantal distance covered= 80.19 m from base of building

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