Imagine a ball fired from a height of 2.0 m at 3.0 m/s, and at 45° above horizontal. If the ball lands on horizontal ground, where does it do so?

1.89 m

Explanation:

h = 2 m

u = 3 m/s

θ = 45

Let t be the time taken

use second equation of motion in vertical direction

h = u t + 1/2 a t^2

- 2 = 3 Sin 45 t - 1/2 x 9.8 x t^2

- 2 = 2.12 t - 4.9 t^2

4.9 t^2 - 2.12 t - 2 = 0

By solving we get

t = 0.89 sec

Let x be the horizontal distance travelled

x = u Cos 45 x t

x = 3 Cos 45 x 0.89 = 1.89 m