A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground and experience negligible air resistance. How far from where the child is standing does the ball hit the ground?

initial velocity, u = 8 m/s

vertical height, h = 1 m

θ = 40°

Let the horizontal distance is d and the time taken is t.

Use second equation of motion in vertical direction

h = ut + 1/2 at²

1 = 8 Sin 40 x t + 0.5 x 9.8 t²

1 = 5.14 t - 4.9t²

4.9t² - 5.14 t + 1 = 0

so, t = 0.26 s (smaller value)

So, the horizontal distance is

d = u cos 40 x t

d = 8 cos 40 x 0.26

d = 1.6 m