Question

# A water balloon is launched at a speed of 27 m/s and an angle of 36...

A water balloon is launched at a speed of 27 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 27 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.

let u = initial speed of balloon = 27m/s

horizontal distance covered by balloon = 27m

by taking eq of motion in horizontal direction:

S_x = u_x*t + (1/2) * a_x*t2 ( here a_x = 0 because velociy is constant in horizontal direction)

27 = u*cos36*t + 0

t = 27/ 27*cos36

t = 1.236 sec

this is the time required by balloon to hit the building.

now taking eq of motion in vertical direction:

S_y = u_y*t + ( 1/2)*a_y*t2

h = u *sin36*t + ( 1/2)*(-g)*t2

h = 27*sin36*1.236 - (1/2)*9.8*(1.236)2

h = 12.1 m

this is the height at which balloon hit the buliding.

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