Question

Suppose a soccer player boots the ball from a distance 33.5 m toward the goal. find the initial speed of the ball if it just passes over the goal,2.4 m above the ground, given the initial direction of 40 degrees above the horizontal.

Answer 1

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the ball = V₀

Direction of initial velocity = $\theta$ = 40^o

Initial horizontal velocity of the ball = V_x0 = V₀Cos$\theta$

Initial vertical velocity of the ball = V_y0 = V₀Sin$\theta$

Horizontal distance of the goal from the initial point = R = 33.5 m

Height of the ball when it crosses the goal = 2.4 m

Time taken by the ball to reach the goal = T

There is no horizontal force acting on the ball therefore the horizontal acceleration is zero.

R = V_x0T

R = (V₀Cos$\theta$)T

$T = \frac{R}{V_{0}Cos\theta }$

H = V_y0T + gT²/2

$H = (V_{0}Sin\theta )(\frac{R}{V_{0}Cos\theta }) + (\frac{g}{2})(\frac{R}{V_{0}Cos\theta })^{2}$

$H = RTan\theta + \frac{gR^{2}}{2V_{0}^{2}Cos^{2}\theta }$

$2.4 = (33.5)Tan(40) + \frac{(-9.81)(33.5)^{2}}{2V_{0}^{2}Cos^{2}(40)}$

$2.4 = 28.11 - \frac{9380.385}{V_{0}^{2}}$

$\frac{9380.385}{V_{0}^{2}} = 25.71$

V₀² = 364.853

V₀ = 19.1 m/s

Initial speed of the ball = 19.1 m/s