A soccer ball is kicked from the ground with an initial speed of 17.8 m/s at an upward angle of 45.9˚. A player 48.6 m 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? Neglect air resistance.
-
First, find the time (t) that the ball will land.
vyf = vyi + a*t
vyf will be the same as the initial speed, but negative. (The ball will land with the same speed that it was kicked with).
vyf = - vyi
-a*t = 2*vyi
t = -2*vyi/a
t = -2*(17.8*sin(45.9˚))/(-9.8m/s²)
t = 3.4163 s
The ball will land x from where it was kicked where:
x = vxi*t
x = (17.8*cos(45.9˚))*(3.4163)
x = 42.32m
If the player is 48.6 m away, then he needs to run (48.6 m - 42.3 m) to get to the ball in 3.4163 seconds.
v = (48.6 - 42.3)/3.4163
v = 1.884 m/s
vyf = vyi + a*t
vyf will be the same as the initial speed, but negative. (The ball will land with the same speed that it was kicked with).
vyf = - vyi
-a*t = 2*vyi
t = -2*vyi/a
t = -2*(17.8*sin(45.9˚))/(-9.8m/s²)
t = 3.4163 s
The ball will land x from where it was kicked where:
x = vxi*t
x = (17.8*cos(45.9˚))*(3.4163)
x = 42.32m
If the player is 48.6 m away, then he needs to run (48.6 m - 42.3 m) to get to the ball in 3.4163 seconds.
v = (48.6 - 42.3)/3.4163
v = 1.884 m/s