Question

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 17.7 m/s at an angle of 31.0˚ 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

vertical journey (accelerated motion)

initial velocity, u= v*sin(31) = 17.7*sin(31) = 9.12 m/s

final velocity, v_y = ?

vertical displacement, s= 3.2 m

acceleration, a = -g = -9.81 m/s²

using, v_y² - u² = 2as

v_y² - 9.12² = -2*9.81*3.2

v_y = 20.4 m/s

Horizontal journey (unaccelerated)

that means the initial horizontal velocity remains unchanged = v*cos(31) = 17.7*cos(31) = 15.17 m/s = v_x

hence final velocity, = sqrt(v_y² + v_x²) = sqrt(20.4² + 15.17²) = 25.422 m/s