Question

A ball is kicked such that it leaves the ground at an upwards angle of 35° above the horizontal with an initial velocity of 17 m s−1.

How far away does the ball hit the ground? (in m to 2.s.f)

(note: ignore the effects of air resistance when solving this problem)

Answer 1

Assume accelerations is $g=9.8\,m\,s^{-2}$ , angle between initial velocity and horizontal line $\theta=35^ \circ$

The distance travelled:

$$d=\frac{v^2 \sin{2\theta}}{g}=\frac{17^2 \sin{70^\circ}}{9.8} m = 27.7 m$$