Asked by Anonymous
A water balloon is launched at a speed of 23 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 20 m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures.
Answers
Answered by
Damon
Horizontal problem:
u = 23 cos 36 until it stops
20 meters = u t
so
t = 20 /(23 cos 36)
use that t in the vertical problem
now vertical problem:
Vi = 23 sin 36 initial speed up
h = 0 + Vi t - 4.9 t^2
u = 23 cos 36 until it stops
20 meters = u t
so
t = 20 /(23 cos 36)
use that t in the vertical problem
now vertical problem:
Vi = 23 sin 36 initial speed up
h = 0 + Vi t - 4.9 t^2
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