Question

A football is kicked with a velocity of 20 m/s at an angle of 37° with the horizontal. Determine:

a. Maximum height reached by the ball

b. Time of flight

c. Horizontal distance traveled

Answer 1

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

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

Then, we can find the maximum height reached by the ball from the kinematic equation:

$$y_{max}=v_0tsin\theta-\dfrac{1}{2}gt^2.$$

Substituting $t$ into the previous equation, we get:

$$y_{max}=v_0sin\theta(\dfrac{v_0sin\theta}{g})-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(20\ \dfrac{m}{s})^2sin^237^{\circ}}{2\times9.8\ \dfrac{m}{s^2}}=7.39\ m.$$

(b) We can find the time of flight as follows:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\times20\ \dfrac{m}{s}\times sin37^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.46\ s.$$

(c) We can find the horizontal distance traveled as follows:

$$x=v_0t_{flight}cos\theta=20\ \dfrac{m}{s}\times2.46\ s\times cos37^{\circ}=39.3\ m.$$