Question

A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff, the stone lands 36.0 meters away. At what speed was the stone launched

Answer 1

V = 333 m/s

Step-by-step explanation:

Given,

The height of the cliff, h = 20 m

The horizontal range of the stone, S = 36 m

The horizontal velocity of the stone, V = ?

The formula for horizontal range is given by,

S = Vx [Vy + √(Vy² + 2gh] / g

Since Vx = 0, the equation becomes

S = Vx√(Vy² + 2gh] / g

Solving for Vx

Vx = Sg - √(2gh)

Sunstituing the given values in the above equation

Vx = 36 x 9.8 - [√(2 x 9.8 x 20)]

= 333 m/s

Hence, the imparted horizontal velocity of the stone is, V = 333 m/s