Question

A rock is projected from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 73° above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. What is the time?

Answer 1

7 s

Step-by-step explanation:

u = Initial velocity of rock = 12.2 m/s

`\theta` = Angle of throw =
`73^(\circ)`

x = Displacement in x direction = 25 m

Displacement in x direction is given by

`x=u\cos\theta t\\\Rightarrow t=(x)/(u\cos\theta)\\\Rightarrow t=(25)/(12.2* \cos73^(\circ))\\\Rightarrow t=7\ \text{s}`

Time taken to reach the ground is 7 s.