Question

A soccer player kicks a ball at an angle of 37 degree from the horizontal with an initial speed

of 50 ft/sec. Assume that the ball moves in a vertical plane. a) Find the time t1 at which the

ball reaches the highest point of its trajectory. b) How high does the ball go? c) What is the

horizontal range of the ball and how long is it in the air? d) What is the velocity of the ball as

it strikes the ground?

Answer 1

a) We can find the time $t_1$ at which the ball reaches the highest point of its trajectory from the kinematic equation:

$$v_y=v_0sin\theta-gt_1,$$ $$0=v_0sin\theta-gt_1,$$ $$t_1=\dfrac{v_0sin\theta}{g}=\dfrac{50\ \dfrac{ft}{s}\cdot sin37^{\circ}}{32\ \dfrac{ft}{s^2}}=0.94\ s.$$

b) We can find the maximum height of the ball from the formula:

$$y_{max}=\dfrac{v_0^2sin^2\theta}{2g}=\dfrac{(50\ \dfrac{ft}{s})^2\cdot sin^237^{\circ}}{2\cdot32\ \dfrac{ft}{s^2}}=14.15\ ft.$$

c) We can find the total flight time of the ball from the formula:

$$t=2t_1=2\cdot0.94\ s=1.88\ s.$$

The horizontal range can be found as follows:

$$x=v_0tcos\theta=50\ \dfrac{ft}{s}\cdot 1.88\ s\cdot cos37^{\circ}=75\ ft.$$

d) The horizontal component of the ball's velocity can be found as follows:

$$v_x=v_0cos\theta=50\ \dfrac{ft}{s}\cdot cos37^{\circ}=40\ \dfrac{ft}{s}.$$

The vertical component of the ball's velocity at time $t=1.88\ s$ can be found as follows:

$$v_y=v_0sin\theta-gt=50\ \dfrac{ft}{s}\cdot sin37^{\circ}-32\ \dfrac{ft}{s^2}\cdot1.88\ s=-30\ \dfrac{ft}{s}.$$

Finally, we can find the velocity of the ball as it strikes the ground from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(40\ \dfrac{ft}{s})^2+(-30\ \dfrac{ft}{s})^2}=50\ \dfrac{ft}{s}.$$

Answer:

a) $t_1=0.94\ s.$

b) $y_{max}=14.15\ ft.$

c) $t=1.88\ s, x=75\ ft.$

d) $v=50\ \dfrac{ft}{s}.$