*41. a soccer player kicks the ball toward a goal that is 16.8 m in front of him. the ball leaves his foot at a speed of 16.0 m/s and an angle of 28.08 above the ground. find the speed of the ball when the goalie catches it in front of the net.

Answer;

V = 14.7 m/s

Explanation and solution;

Vx = V cos theta  

Vx = 16m/s × cos 28.08  

Vx = 14.1 m/s  

Vy = Vsin theta  

Vy = 7.51m/s  

t = d/Vx  

t = 16.8m/ 14.1  

t= 1.191 s  

Now, find Vy at t= 1.191  

Vy = Vy - gt  

Vy = 7.51 - 9.81*1.191  

Vy = - 4.18 m/s  

V = sqrt (Vy^2 +Vx^2)  

V= sqrt (4.18^2 + 14.1^2)  

V= 14.7m/s

A

R2 = Rx2 + Ry2

Explanation:

To find the magnitude of resultant vector or resultant forces, we must first resolve the vectors into X component and Y component. That is,

The summation of all the vectors in the X direction will be Rx while

The summation of all the vectors in the Y direction will be Ry

The magnitude of the resultant vector R will therefore,

R^2 = Rx^2 + Ry^2