Question

Suppose the body in Sample Problem 16.2c is projected at 24.5 m/s, 60º above the horizontal. Find the:

(a) Time of flight

(b) Range

Answer 1

(a) Let's first find the time that the body takes to reach its maximum height:

$$v_y=v_0sin\theta-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

The time of flight can be found as follows:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\cdot24.5\ \dfrac{m}{s}\cdot sin60^{\circ}}{9.8\ \dfrac{m}{s^2}}=4.33\ s.$$

(b) We can find the range as follows:

$$x=v_0t_{flight}cos\theta,$$ $$x=v_0cos\theta\cdot\dfrac{2v_0sin\theta}{g}=\dfrac{v_0^2sin2\theta}{g},$$ $$x=\dfrac{(24.5\ \dfrac{m}{s})^2sin2\cdot60^{\circ}}{9.8\ \dfrac{m}{s^2}}=53.04\ m.$$