Question

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.

Answer #1

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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