A man stands on the roof of a building of height 16.8 m and throws a rock with a

Question

A man stands on the roof of a building of height 16.8m and throws a rock with a velocity of magnitude 31.8m/s at an angle of 25.0? above the horizontal. You can ignore air resistance.

part A: Calculate the maximum height above the roof reached by the rock. I was able to find this which was 9.21 m

part B: Calculate the magnitude of the velocity of the rock just before it strikes the ground?

part C: Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Explanation / Answer

vertical component of speed = vsin(theta) = 31.8*sin25 = 13.43

v^2 = u^2 - 2*9.8*s

0 = 13.34^2 -2*9.8*s

s = 9.21 m

B)total height above ground = 9.21+ 16.8 = 26

free fall velocity from that height

V^2 = u^2 + 2*9.8*26

v(vertical component) = 22.57

horizontal component(unchanged) = 31.8*cos25 = 28.82 m/sec

resultant = sqrt(22.57^2 + 28.82^2) = 36.6 m/sec

C)time taken for particle to reach the top

v = u - 9.8*t1

0 = 13.34 -9.8*t1

t1= 1.36 sec

time taken to drop from height of 26m

h = ut + 0.5*9.8*t2^2

26 = 0 + 0.5*9.8*t2^2

t2 = 2.3 sec

total time = 3.66 sec

thus horizontal distance covered = 28.82*3.66 = 105.58 m

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