Question

A student drops a ball form a window 3.5 meters above the sidewalk, How fast is it moving when it hits the sidewalk

Answer 1

Answer on Question #44882, Physics, Mechanics | Kinematics | Dynamics

Question:

A student drops a ball form a window 3.5 meters above the sidewalk. How fast is it moving when it hits the sidewalk

Answer:

The law of conservation of energy:

$$T + U = \text{const}$$

where $T = \frac{mv^2}{2}$ is kinetic energy, $m$ - mass of the body, $v$ - speed

$U = mgh$ is potential energy, $g$ - gravitational acceleration, $h$ - height

$$T_1 + U_1 = T_2 + U_2$$

1 - initial state: $T_1 = 0, U_1 = mgh$

2 - final state: $T_2 = \frac{mv^2}{2}, U_2 = 0$

Therefore:

$$mgh = \frac{mv^2}{2}$$ $$v = \sqrt{2gh} \cong 8.3 \frac{m}{s}$$

Answer: $8.3 \frac{m}{s}$

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