Question

A projectile is shot from the edge of a vertical cliff 60.0 m above the ocean. It has a speed of 100 m/s and is fired at an angle of 35.0° below the horizontal. How far from the foot of the vertical cliff does the projectile hit the water?

Answer 1

Answer:

79.5 m

Explanation:

Let t be the time taken to hit the surface of water and x be the horizontal distance traveled.

use II equation of motion in Y axis direction

h = uy t + 1/2 g t^2

- 60 = - 100 Sin 35 x t - 1/2 x 9.8 x t^2

-60 = - 57.35 t - 4.9 t^2

4.9 t^2 + 57.35 t - 60 = 0

$t = \frac{-57.35\pm \sqrt{57.35^{2} + 4 \times 4.9 \times 60}}{2\times 4.9}$

By solving we get

t = 0.97 second

The horizontal distance traveled is

x = ux t

x = 100 Cos 35 x 0.97

x = 79.5 m