A soccer player kicks the ball toward a goal that is 17.0 m in front of him. The ball leaves his foot at a speed of 16.0 m/s and an angle of 31.0 ° above the ground. Find the speed of the ball when the goalie catches it in front of the net.

Question

anonymous · Answer

v=16 , θ=31 , x = 17m
Vx,the horizontal component of velocity = v*cos θ
Voy,the initial vertical component of velocity = v*sin θ
Vx will remain unchanged throughout the motion of the ball because there is no force in the horizontal direction. Motion in the vertical direction will be acted with gravity and the acceleration in the vertical direction is "g" and hence Voy will continuously change.
Time taken by ball to reach the goal = t = x/Vx

Vertical velocity of the ball after time t can be found by :
Vy=Voy -gt     where t=x/Vx
The magnitude of velocity(speed) is then given by

v =√(Vx)^2+(Vy)^2