Question

a ball of mass 50 g dropped from height of 20 m . a boy on the ground hits the ball vertically upwards with a bat with an average force of 200N so that it attains a vertical height of 45 m .the time for which the ball remains in contact with the bat g=10 m/s square

Answer 1

A ball of mass $50\,\mathrm{g}$ dropped from height of $20\,\mathrm{m}$, a boy on the ground hits the ball vertically upwards with a bat with an average force of $200\,\mathrm{N}$ so that it attains a vertical height of $45\,\mathrm{m}$. The time for which the ball remains in contact with the bat $\mathrm{g} = 10\,\mathrm{m/s}$ square

Solution

According to the conservation of energy law potential energy of the ball on height $h_1 = 20\,\mathrm{m}$ is equal to kinetic energy of the ball on the ground:

$$m g h _ {1} = \frac {m v _ {1} ^ {2}}{2} \rightarrow v _ {1} = \sqrt {2 g h _ {1}}.$$

When a boy on the ground hits the ball vertically upwards it have kinetic energy which equal potential energy of the ball on height $h_1 = 45\,\mathrm{m}$:

$$\frac {m v _ {2} ^ {2}}{2} = m g h _ {2} \rightarrow v _ {2} = \sqrt {2 g h _ {2}}.$$

Impulse transmitted to the ball by bat:

$$I = F \Delta t = P _ {2} - P _ {1} = m v _ {2} - (- m v _ {1}) = m (v _ {1} + v _ {2}) = m \left(\sqrt {2 g h _ {1}} + \sqrt {2 g h _ {2}}\right),$$

$P_{1}$ is negative because it is opposite to direction $f$ force $F$.

A time for which the ball remains in contact with the bat:

$$\begin{array}{l} \Delta t = \frac {m \left(\sqrt {2 g h _ {1}} + \sqrt {2 g h _ {2}}\right)}{F} = \frac {5 0 * 1 0 ^ {- 3} \mathrm {k g} \left(\sqrt {2 * 1 0 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} * 4 5 \mathrm {m}} + \sqrt {2 * 1 0 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} * 2 0 \mathrm {m}}\right)}{2 0 0 \mathrm {N}} \\ = 0. 0 1 2 5 \mathrm {s}. \end{array}$$

Answer: 0.0125s.