Question

A golfer hits a shot to a green that is elevated 2.60 m above the point where the ball is struck. The ball leaves the club at a speed of 19.2 m/s at an angle of 48.0˚ 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

initia components
voy = vo*sin(48) = 19.2*sin(48) = 14.27 m/s

vox = vo*cos(48) = 19.2*cos(48) = 12.85 m/s

let t is the time taken for the journey.

Apply, h = voy*t - 0.5*g*t^2

2.6 = 14.27*t - 4.9*t^2

4.9*t^2 - 14.27*t + 2.6 = 0

on solving the above equation we get

t = 2.717 s

so,

when the the ball stuck

vy = voy - g*t

= 14.27 - 9.8*2.717

= -12.36 m/s

vx = vox = 12.85 m/s

v = sqrt(vx^2 + vy^2)

= sqrt(12.85^2 + 12.36^2)

= 17.8 m/s <<<<<<--------------Answer