Answer to Question #175009 in Classical Mechanics for Mary Chumanni

Question

A student throws a water balloon vertically downward from the top of the building. The balloon leaves the thrower’s hand with a speed of 6.0m/s. Air resistance may be ignored so the water balloon is in free fall after it leaves the thrower’s hand.

A.) What is it’s speed after falling 2.0s?

B.) How far does it fall in 2s?

C.) What is the magnitude of it’s velocity after falling 10m?

Answer 1

Answer

Using newton equation of motion

A. Velocity after t time

v=u+gt

$v=0+9.8\times2=19.6m/s$

B. Distance travelled after t=2sec

$S=ut+0.5gt^2\\=0+0.5\times9.8\times4\\=19.6m$

C. Velocity after t=10sec

$v=0+9.8\times10=98m/s$