Question

2. An object is projected 60 ⁰ from the ground with an initial velocity of 24.5 m/s. Find the (a) Total time of Flight (b) Range (c) Maximum height

Answer 1

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

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

Then, the total 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 of the object from the kinematic equation:

$$x=v_0t_{flight}cos\theta=24.5\ \dfrac{m}{s}\cdot4.33\ s\cdot cos60^{\circ}=53.04\ m.$$

(c) We can find the maximum height attained by the object from the kinematic equation:

$$y_{max}=v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(24.5\ \dfrac{m}{s})^2sin^260^{\circ}}{2\cdot9.8\ \dfrac{m}{s^2}}=22.97\ m.$$