Answer to Question #94082 in Physics for cynthiaafria

Question

A soccer ball is kicked from the ground with any initial speed of 19.5 m/s at an upward angle of 45 degree. A player 55 m away on the direction of the kick start running to meet the ball at that instant. What just be his average speed if he is to meet the ball just before it hits the ground.

Answer 1

Let us use notation: $L$ - horizontal distance, covered by the ball until it reaches the ground, $S = 55 m$ - the distance from the ball to the soccer, $v_0 = 19.5 \frac{m}{s}$ - initial speed of the ball, $\theta = 45^{\circ}$ - the angle of the ball.

In order for the player to meet the ball just before it hits the ground, he/she must cover the distance $\Delta = S - L$ in time $T$, during which the ball covers the distance $L$ . Assuming that the player is moving with the constant speed, it has to be equal to $v = \frac{\Delta}{T} = \frac{S - L}{T}$ .

According to the formulas of projectile motion, $T = \frac{2 v_0 \sin \theta}{g}$ , $L = \frac{v_0^2 \sin 2\theta}{g}$ , therefore $v = \frac{S - L}{T} = \frac{S g}{2 v_0 \sin \theta} - \frac{v_0 \cos \theta}{2} \approx 12.67 \frac{m}{s}$ .