A person wants to launch a water balloon and hit the top of a tower, 50.0 meters away. The water balloon launcher is angled at 25 degrees. The water balloon leaves the launcher at 68 m/s. If the balloon hits the top of the tower, how high was the tower?
vertical problem:
Vi = 68 sin 25 = 28.7 m/s
h = Vi t -4.9 t^2
h = 28.7 t - 4.9 t^2
horizontal problem:
u = 68 cos 25 = 61.6 m/s
t = 50/61.6 = .811 s
so
h = 28.7(.811) - 4.9(.811^2)
convert vector from polar to rectangular form
horiz: 61.63 m/s
vert: 28.74 m/s
t = 50/61.63 = 0.81 s to impact
what is the height of the balloon after that time?
Damon, I understand finding the vertical and horizontal velocities.I start getting confused when I have to integrate the two velocities and use time to find height
could you break down this problem a little more and explain what is happening?
I think of it this way:
you have a velocity vector at an elevation of 25 degrees and a magnitude of 68
the base (start point) of the vector is at ground level, i.e., the launcher is not elevated
the two velocities (H and V) coexist upon firing
the H velocity is constant
the V velocity changes due to gravity, so the balloon goes up until V velocity becomes zero, then it increases as the balloon comes down
the timing is such that at the same instant the H velocity causes the balloon to reach the tower, the balloon has also fallen to the same height as the tower and makes contact
does that help?