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a soccer ball is kicked from the ground with an initial speed of 19.5 m/s at an upward angle of 45o. a player 55m away in the direction of the kick starts running to meet the ball at that instant. what must be his average speed if he is to meet the ball just before it hits the ground?



Answer :

Answer:

R = V^2 sin 2θ / g        range formula

R = 19.5^2 * 1 / 9.80 = 38.8 m

Vx = 19.5 cos 45 = 13.8 m/s       horizontal speed

t = R / Vx = 38.8 / 13.8 = 2.81 sec

v = (55 - 38.8) / 2.81 = 5.76 m/s   (about 12.9 mph)

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