Question

The roof of a two-story house makes an angle of 29° with the horizontal. A ball rolling down the roof rolls off the edge at a speed of 4.9 m/s. The distance to the ground from that point is 6.4 m. (a) How long is the ball in the air? s (b) How far from the base of the house does it land? m (c) What is its velocity just before landing? (Let upward be the positive y-direction.) x-component m/s y-component m/s

Answer 1

`t = 0.93 s`

Part b)

`d = 3.98 m`

Part c)

`v_x = 4.28 m/s`

`v_y = -11.49 m/s`

Step-by-step explanation:

The two components of the velocity of the ball is given as

`v_x = 4.9 cos29`

`v_x = 4.28 m/s`

`v_y = 4.9 sin29`

`v_y = 2.37 m/s`

Part a)

now we know that the displacement in y direction is given as

`\Delta y = 6.4 m`

so we have

`\Delta y = v_y t + (1)/(2)gt^2`

`6.4 = 2.37 t + 4.9 t^2`

`t = 0.93 s`

Part b)

Distance of the ball in x direction of the motion is given as

`d = v_x t`

`d = 4.28 * 0.93`

`d = 3.98 m`

Part c)

In x direction the velocity will remain the same always

`v_x = 4.28 m/s`

while in Y direction we can use kinematics

`v_y = v_(oy) + at`

`v_y = -2.37 - 9.81(0.93)`

`v_y = -11.49 m/s`