Question

a soccer player running on a level playing field kicks a soccer ball with a velocity of 15 m/s at an angle of 60° above the horizontal. determine the soccer ball’s (a) time of flight (b) range (c) maximum height

Explanation:

Step1: Find the initial vertical velocity

The initial velocity \(v_0 = 15\ m/s\) and the angle \(\theta=60^{\circ}\). The initial vertical velocity \(v_{0y}=v_0\sin\theta\).

$$v_{0y}=15\times\sin60^{\circ}=15\times\frac{\sqrt{3}}{2}\approx12.99\ m/s$$

Step2: Calculate the time of flight (a)

The time of flight \(T\) for a projectile motion (when the ball returns to the same - height level) is given by the formula \(T = \frac{2v_{0y}}{g}\), where \(g = 9.8\ m/s^{2}\).

$$T=\frac{2\times12.99}{9.8}\approx2.65\ s$$

Step3: Calculate the range (b)

The initial horizontal velocity \(v_{0x}=v_0\cos\theta\). \(v_{0x}=15\times\cos60^{\circ}=15\times\frac{1}{2}=7.5\ m/s\).
The range \(R\) is given by \(R = v_{0x}\times T\).

$$R=7.5\times2.65 = 19.875\ m$$

Step4: Calculate the maximum height (c)

The maximum - height formula is \(H=\frac{v_{0y}^{2}}{2g}\).

$$H=\frac{(12.99)^{2}}{2\times9.8}=\frac{168.74}{19.6}\approx8.61\ m$$

Answer:

(a) The time of flight is approximately \(2.65\ s\).
(b) The range is approximately \(19.88\ m\).
(c) The maximum height is approximately \(8.61\ m\).