You stand on your balcony 10m above the ground. You throw a water balloon at a friend who is standing 5m from the base of the building.
If it hits him 0.5s later, what was the velocity with which you threw the balloon and what was its velocity an instant before it hit him?
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A balloon is thrown from a balcony 10 m above the ground at a person standing 5 m from the base of the building. Let the velocity at which it is thrown be V in a direction of x degrees made with the vertical.
The vertical component of the velocity is V*cos X. The balloon is accelerated vertically downwards at 9.8 m/s^2. The time taken by the balloon to travel 10 m is 0.5 s which gives 10 = V*cos X*0.5 + (1/2)*9.8*(.5)^2
=> V*cos X = 17.55 ...(1)
There is no horizontal acceleration acting on the balloon. The horizontal component of the velocity is V*sin X. This gives 5 = V*sin X*0.5
=> V*sin X = 10 ...(2)
(2)/(1) = tan X = 10/17.55
=> X = 29.67 degrees
V = sqrt(17.55^2 + 100) = 20.19 m/s
The balloon is thrown at 20.19 m/s in a direction 29.67 degrees with the vertical direction. When the balloon hits the person below, its velocity is sqrt(10^2 + (17.55 + 4.9)^2) = 24.57 m/s and it makes an angle equal to arc tan(10/(17.55 + 4.9)) = 24 degrees.
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