Question #107658

A 0.5 kg water balloon is dropped from a balcony 40 meters from the ground. What is the balloon's velocity just as it hits the ground? (Ignoring air resistance)

Expert's answer

Ignoring the air resistance means the case of free fall (gravity is the only force acting upon body).

h=v0t+gt22h=v_0 t+\frac{gt^2}{2}

and

v=v0+gtv=v_0+gt,

where hh - height, m; vv - velocity, m/s; v0v_0 - initial velocity, m/s; tt - time, s; gg - acceleration due to gravity, m/s2 (g9.8  m/s2g \approx 9.8 \; m/s^2). In our case v0=0v_0 =0, so balloon will hit the ground at

t=2hgt= \sqrt{\frac{2h}{g}}

v=g2hg=2ghv=g \cdot \sqrt{\frac{2h}{g}} = \sqrt{2gh}

v=29.840=784=28  m/sv=\sqrt{2 \cdot 9.8 \cdot 40} = \sqrt{784} = 28 \; m/s


Answer: 28 m/s.


LATEST TUTORIALS
APPROVED BY CLIENTS