a ball is thrown up onto a roof, landing 4.70 s later at height h = 24.0 m
above the release level. The ball's path just before landing is angled at � = 62.0° with the roof.
(a) Find the horizontal distance d it travels. (Hint: One way is to reverse the motion, as if it is
on a video.) What are the (b) magnitude and (c) angle (relative to the horizontal) of the ball's
initial velocity?
To find the horizontal distance the ball travels, we can use the equation:
d = v₀ * t
where:
d = horizontal distance traveled
v₀ = initial horizontal velocity
t = time of flight
To find the initial horizontal velocity, we need to separate the initial velocity into horizontal and vertical components. The horizontal component of the velocity remains constant, while the vertical component changes due to the effect of gravity.
To find the initial velocity, we can use the equation:
v₀ = v * cos(θ)
where:
v₀ = initial velocity
v = magnitude of initial velocity
θ = angle of the ball's initial velocity relative to the horizontal
Now, let's find the initial velocity by using the given information.
(a) Finding the horizontal distance traveled:
The time of flight, t, is given as 4.70 seconds.
So, we can use the equation d = v₀ * t to find the horizontal distance.
(b) Finding the magnitude of the ball's initial velocity:
The height, h, is given as 24.0 m. The vertical distance traveled can be found using the equation:
h = v₀y * t - 0.5 * g * t²
Since the ball is thrown up, the initial vertical velocity, v₀y, is positive, and the acceleration due to gravity, g, is negative.
Now, let's calculate the magnitude of the initial velocity:
Using the provided information, we have:
h = 24.0 m and t = 4.7 s
(c) Finding the angle of the ball's initial velocity:
The given angle of the ball's initial velocity, θ, is 62.0° with the roof.
With this information, we have found the horizontal distance traveled (part a), magnitude of the ball's initial velocity (part b), and the angle of the ball's initial velocity relative to the horizontal (part c).