Question

A man stands on the roof of a 10.0 m -tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 42.0 ∘ 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

Answer 1

A) vertical component of speed = vsin(theta)

                                             = 30*sin42 = 20.07

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

0 = 20.07^2 - 2*9.8*s

s = 20.56 m

B) total height above ground = 20.56 + 10 = 30.56

free fall velocity from that height

v^2 = u^2 + 2*9.8*30.56

v^2 = 0 + 2*9.8*30.56

v(vertical component) = 24.47

horizontal component(unchanged) = 30*cos42 = 22.3 m/sec

resultant = sqrt(24.47^2 + 22.3^2)

            = 33.1 m/sec

C)time taken for particle to reach the top

v = u - 9.8*t1

0 = 20.07 - 9.8*t1

t1= 2.04 sec

time taken to drop from height of 30.56 m

h = ut + 0.5*9.8*t2^2

30.56 = 0 + 0.5*9.8*t2^2

t2 = 2.5 sec

total time = 4.54 sec

thus horizontal distance covered = 22.3*4.54

                                               = 101.2 m