Question

a ball is thrown up onto a roof, landing 3.60 s later at height h = 25.0 m above the release level. The ball's path just before landing is angled at ? = 64.0?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?

Answer 1

V = initial velocity of the ball
H = maximum height reached by the ball
t = time to reach max height
T = time the ball reaches the roof = 3.6
T - t = time it takes the ball to fall from max height to the roof
H - 25 = distance the ball falls from max height to the roof
g = acceleration of gravity = 9.81 m/s^2

Use the equations for calculating max height and for the ball to fall from max height to the roof.
H = (1/2)*g*t^2
H - 25 = (1/2*g*(T - t)^2 = (1/2*)g*(3.6 - t)^2

Put the first equation into the second for H.
(1/2)*g*t^2 - 25 = (1/2)*g*(3.6 - t)^2 = (1/2)*g*[12.96 - 7.2*t + t^2]
-25 = (1/2)*g*(12.96 - 7.2*t) = 4.905*(12.96 - 7.2*t)
-5.097 = 12.96 - 7.2*t
7.2*t = 18.057
t = 2.507 seconds

Now use this time to get the initial vertical velocity.
v = g*t = 9.81*2.507 = 24.602 m/s

Use the vertical impact velocity and angle to get the horizontal component of velocity at impact. This will equal the initial horizontal velocity since there are no forces acting in the horizontal.
Time to fall = 3.6 - 2.507 = 1.093 seconds
Vertical velocity = v = g*(time to fall) = 9.81*1.093 = 10.722 m/s

tan(64) = (vertical velocity)/(horizontal velocity)
Horizontal velocity = 10.722/tan(50) = 8.99 m/s

The initial velocity is then:
V = SQRT[(8.99)^2 + (24.602)^2]
V = 26.193 m/s

Angle is:
tan(Angle) = (vertical velocity)/(horizontal velocity) = 24.602/8.99
tan(Angle) = 2.7365
Angle = 69.9268 degrees

The initial velocity is 26.193 m/s and the angle with the horizontal is 69.9268 degrees.

. Distance is just the horizontal speed times the time of flight.
Distance = 8.99*3.6
Distance = 32.364 meters

So the ball traveled a horizontal distance of 24.602 meters