Question

A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stop watch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff? (g = 9.8 m/s² and air resistance is negligible)

Answer 1

the height was thrown = 120 m

Step-by-step explanation:

given,

speed of the rock = 12 m/s

angle of launch = 30.0 ∘

horizontal distance = d = 15.5 m

acceleration due to gravity = 9.8 m/s²

`u_y = u sin \theta`

`u_y = 12 sin 30^0`

`u_y = 6 m/s`

t =5.6

`s_y = u_y t + (1)/(2)gt^2`

`s_y = 6* 5.6 - (1)/(2)* 9.8 * 5.6^2`

`s_y = -120.064m`

hence, the height was thrown = 120 m