Question

A rock is projected from the edge of the top of a building with an initial velocity of 12 m/s at an angle of 37? above the horizontal. The rock strikes the ground ahorizontal distance of 35 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

Answer 1

Let the height of the building be h

Now, initial horizontal velocity of ball = 12*cos37 = 12*0.8 = 9.6 m/s

Now, time taken by ball to reach the ground = 35/9.6 sec

But time of flight while in projectile motion = 2*12*sin37/9.8 = 1.47 sec

Also, initial velocity of ball in vertically downward direction = 12*sin37 = 12*0.6 = 7.2 m/s

So,

Net time taken by ball to cover vertical distance of h = 35/9.6 - 1.47 = 3.64 - 1.47 = 2.17 sec

Applying the kinematic equations, we have,

h = ut + 0.5*g*t²

= 7.2*2.17 + 0.5*9.8*2.17²

= 15.62 + 23.07

= 38.69 m