Question

A 0.86 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.

Answer 1

Step-by-step explanation:

Given

mass of rock=0.86 kg

initial velocity=8.65 m/s

launch angle
`=46^(\circ)`

horizontal distance=13.7 m

let
`u_x` be the horizontal velocity given by
`8.65\cos 46`

and
`u_y` be the vertical velocity
`= 8.65\sin 46`

`R=u_x* t`

`13.7=8.65\cos 46* t`

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

`T=(2u\sin \theta )/(g)`

`T=(2* 8.65* \sin 46)/(9.8)=1.269 s`

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

Let h be the height

`h=u\sin \theta t+(gt^2)/(2)`

h=6.27+4.98=11.25 m

Answer 2

Height of the building = 11.4 m

Step-by-step 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

`v_x = 8.65 cos46`

`v_x = 6 m/s`

vertical component of the speed is given as

`v_y = 8.65 sin46`

`v_y = 6.22 m/s`

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

so we will have

`x = v_x t`

`13.7 = 6 t`

`t = 2.28 s`

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

`y = v_y t + (1)/(2)at^2`

`y = 6.22(2.28) - (1)/(2)(9.81)(2.28^2)`

`y = -11.4 m`

Height of the building = 11.4 m