A golfer hits a shot to a green that is elevated 3.30 m above the point where the ball is struck. The ball leaves the club at a speed of 19.3 m/s at an angle of 35.0˚ above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
initial velocity = u = 19.3 m/sec
horizontal component of u = ux = ucos 35 = 15.80 m/sec
Vertical component of u = uy = usin35 = 11.07 m/sec
Vertical distance = y =3.30 m
Acceleration in y direction = ay = -g
If we ignore the air resistance, the horizontal component of velocity will remain constant. hence there is no acceleration along X-direction. Thus, ax = 0
The final velocity along x-direction = vx = ux+axt = ux =15.80 m/sec
The final velocity along y -direction = vy
= 57.8649
vy = 7.60 m/sec
the final speed
The velocity of ball just before land will be 17.53 m/sec
Pls check the calculations
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