Question

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?

Answer 1

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)

Answer 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?

Answer 3

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

Answer 4

could you break down this problem a little more and explain what is happening?

Answer 5

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?

Answer 6

Yes, it does thank you!

Answer 7

you're quite welcome