Your want to hit your friend with water balloons from the roof of a building on

Question

Your want to hit your friend with water balloons from the roof of a building on a planet with 4.4 x the mass of Earth and 3.7 x the radius of Earth which orbits a 2E+28 kg star. Your friend is 5 feet 6.93 inches tall and stands 12m horizontally away from the building whose the roof is 38 feet 4.63 inches from the ground. Assume the gravitational field strength of Earth is 9.8 N/kg. Your friend makes a 3m/s run back to the school entrance 12m away to find shelter. The entrance is directly below you. How many seconds, after your friend starts running, should you drop the balloon to hit your friend in the head on the way into the building? Answer is 1.5 s

Explanation / Answer

acceleration due to gravity g = G*Mp/Rp^2

Mp = 4.4*ME

Rp = 3.7*RE

g = G*4.4*ME/(3.7*RE)^2

g = 4.4/3.7^2*(G*ME/RE^2)

g = 0.32*gE = 0.32*32

g = 10.24 ft/s^2

===============

1 inch = 0.0833 feet

along horizontal

time T = x/v = 12/3 = 4 s

height of the floor y0 = 38 feet 4.63 inches = 38 + (4.63*0.0833) = 38.4 ft

height of your friend y = 5 feet 6.93 inches = 5 + (6.93*0.0833) = 5.58 ft

from equation of motion

y - y0 = voy*t + (1/2)*ay*t^2

ay = -10.24 ft/s

5.58 - 38.4 = 0 - (1/2)*10.24*t^2

time taken to fall on your friend t = 2.5 s

time after after which you should drop = T - t = 4 - 2.5 = 1.5 s <<<<-----------ANSWER

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