Question

Two students are on a balcony 18.4m above the street. One student throws a ball vertically downward at 15.3 m/s. what is the velocity of the ball as it strikes the ground?

Answer 1

Two students are on a balcony 18.4m above the street. One student throws a ball vertically downward at 15.3 m/s. what is the velocity of the ball as it strikes the ground?

The law of conservation of energy:

$T + U = const$ - the sum of potential energy (PE) and kinetic energy (KE) is the total mechanical energy is constant:

$$T = \frac {m v ^ {2}}{2} - \text{kinetic energy}$$

m - mass of the body

v - speed

$$U = mgh - \text{potential energy}$$

g - gravitational acceleration

h - high

$$T _ {1} + U _ {1} = T _ {2} + U _ {2}$$

1 - initial state: $T_{1} = \frac{mv_{0}^{2}}{2}, U_{1} = mgh$

2 - final state: $T_{2} = \frac{mv^{2}}{2}, U_{2} = 0$

Therefore:

$$\frac {m v ^ {2}}{2} = \frac {m v _ {0} ^ {2}}{2} + mgh$$ $$v = \sqrt {v _ {0} ^ {2} + 2 g h} = \sqrt {1 5 . 3 ^ {2} + 2 * 9 . 8 * 1 8 . 4} = 2 4. 4 \frac {m}{s}$$

Answer: $24.4\frac{m}{s}$