Question

A golfer, standing on a fairway, hits a shot to a green that is elevated 4.00 m above the point where she is standing. The ball leaves her club at an angle of 38.0° above the ground, with a speed of 25.0 m/s.
Find the time that the ball is in the air before it hits the green.

Tries 0/8

How far did the ball travel horizontally?

Tries 0/8

Determine the ball's speed right before it hits the green.

Answer 1

let
h = 4 m
theta = 38 degrees
vo = 25 m/s

vox = vo*cos(theta) = 25*cos(38) = 19.7 m/s
voy = vo*sin(theta) = 25*sin(38) = 15.4 m/s

a) let t is the time taken for the ball to hit the green.

use, h = voy*t - (1/2)*g*t^2

4 = 15.4*t - (1/2)*9.8*t^2

==> t = 2.86 s <<<<<--------------Answer

b) Horizontal distance travelled by the ball,
x = vox*t

= 19.7*2.86

= 56.3 m <<<<<--------------Answer

c) when the ball hits the green,

x-component of velocity, vx = vox = 19.7 m/s

y-component of velocity, vy = voy - g*t

= 15.4 - 9.8*2.86

= -12.6 m/s

speed of the ball when it hits the ground,

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

= sqrt(19.7^2 + 12.6^2)

= 23.4 m/s <<<<<--------------Answer