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 4.5 m/s. The distance to the ground from that point is 8.2 m.
(a) How long is the ball in the air?
(b) How far from the base of the house does it land?
(c) What is its velocity just before landing? (Let upward be the positive y-direction.)
Now I did the squash one. Based on that you should be able to do this one.
To solve this problem, we can break it down into three parts: finding the time of flight, the horizontal distance traveled, and the final velocity before landing.
(a) To find the time of flight, we need to consider the vertical motion of the ball. Since the ball rolls off the edge of the roof, it has an initial vertical velocity of 0 m/s. We can use the equation for vertical motion:
h = (1/2) * g * t^2
Where:
h = vertical displacement (8.2 m)
g = acceleration due to gravity (-9.8 m/s^2)
t = time of flight (unknown)
Rearranging the equation, we get:
t = sqrt((2h) / g)
Substituting the values, we have:
t = sqrt((2 * 8.2) / 9.8) = sqrt(16.4 / 9.8) = sqrt(1.6735) ≈ 1.29 seconds
So, the ball is in the air for approximately 1.29 seconds.
(b) To find the horizontal distance traveled, we can use the equation for horizontal motion:
d = Vx * t
Where:
d = horizontal distance traveled (unknown)
Vx = horizontal component of velocity (initial velocity of the ball)
The horizontal component of the initial velocity can be found using trigonometry:
Vx = V * cos(θ)
Where:
V = initial velocity of the ball (4.5 m/s)
θ = angle of the roof (29°)
Substituting the values, we have:
Vx = 4.5 * cos(29°) ≈ 3.97 m/s
Using this value for Vx and the time from part (a), we can find the horizontal distance traveled:
d = (3.97 m/s) * (1.29 s) ≈ 5.11 meters
So, the ball lands approximately 5.11 meters from the base of the house.
(c) To find the final velocity before landing, we need to consider both the horizontal and vertical components of the velocity. The horizontal component remains constant at 3.97 m/s, while the vertical component changes due to the force of gravity.
The final vertical velocity just before landing can be found using the equation:
Vfy = Vy + g * t
Where:
Vfy = final vertical velocity (unknown)
Vy = vertical component of initial velocity (0 m/s)
Substituting the values, we have:
Vfy = 0 + (-9.8 m/s^2) * (1.29 s) ≈ -12.65 m/s
The negative sign indicates that the velocity is directed downward.
Thus, the final velocity just before landing is approximately 3.97 m/s horizontally and -12.65 m/s vertically.