A0.86 kg rock is projected from the edge of the top of a building with an initial velocity of 8.65 m/s at an angle 46◦ above the horizontal. due to gravity, the rock strikes the ground at a horizontal distance of 13.7 m from the base of the building. building 13.7 m 46◦ 8.65 m/s h how tall is the building? assume the ground is level and that the side of the building is vertical. the acceleration of gravity is 9.8 m/s2 . answer in units of m.

Height of the building = 11.4 m

Explanation:

As we know that the stone is projected at an angle 46 degree with speed 8.65 m/s

so the two components of the speed is given as

vertical component of the speed is given as

now we know that the ball strike at horizontal distance of 13.7 m

so we will have

now we know that in vertical direction ball will move under uniform gravity so we can use kinematics

Height of the building = 11.4 m

Explanation:

Given

mass of rock=0.86 kg

initial velocity=8.65 m/s

launch angle

horizontal distance=13.7 m

let be the horizontal velocity given by

and be the vertical velocity

t=2.279 s

and rock will start covering building height after the completion of its Projectile motion

time taken by stone for zero vertical displacement

thus time taken by rock to cover building height is 2.279-1.269=1.009 s

Let h be the height

h=6.27+4.98=11.25 m