Question

A \[0.5kg\] rubber ball is thrown straight up into the air. At a height of \[20m\] above the ground, it is travelling at \[15m/s\] . How do you answer the following questions?
(A) What is the ball's kinetic energy at this point?
(B) What is its gravitational potential energy relative to the ground?
(C) If all of the \[{E_g}\] the ball has in part b was originally \[{E_k}\] at the ground, calculate how fast the ball was travelling initially when it was thrown up in the air.
(D) Determine the maximum height and velocity reached by the ball (assuming it started at ground level).

Answer 1

Hint: The question asks four quantities. We can start by finding the kinetic energy of the given motion by using the formula, then we can move on to finding the potential energy. Once we have found the values of both of these, we can find the initial velocity by equating the kinetic energy and potential.

Formulas used: The kinetic energy of the system is given by the formula \[{E_k} = \dfrac{1}{2}m{v^2}\]
The potential energy of the system is given by the formula \[{E_g} = mgh\]

Complete step by step solution:
We can start by writing down the values given in the question
The velocity of motion will be \[v = 15m/s\]
The mass of the weight is given as \[m = 0.5kg\]
The height is given as \[h = 20m\]
Now that we have written down the given values, we can find the quantities one by one
(A) The kinetic energy of the ball can be found using the formula \[{E_k} = \dfrac{1}{2}m{v^2}\]
Substituting the values, \[{E_k} = \dfrac{1}{2}m{v^2} = \dfrac{1}{2} \times 0.5 \times {15^2} = 56.25J\]
(B) The potential energy of the ball can be found using the formula, \[{E_g} = mgh\]
Substituting the values, \[{E_g} = mgh = 0.5 \times 9.8 \times 20 = 98J\]
(C) We can find the initial velocity with which the body moves by equating the kinetic energy to the potential energy; that is \[\dfrac{1}{2}m{v^2} = mgh\]
We can cancel the same terms in either side and get \[\dfrac{1}{2}{v^2} = gh\]
Now we can find the velocity using the formula \[v = \sqrt {2gh} \]
Substituting the values, we get \[v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 20} = 19.79m/s\]
(D) The formula to find the maximum height till which the ball goes is \[{H_{\max }} = \dfrac{{{u^2}}}{{2g}} + \dfrac{{{v^2}}}{{2g}}\]
Substituting the values, we get \[{H_{\max }} = \dfrac{{{u^2}}}{{2g}} + \dfrac{{{v^2}}}{{2g}} = \dfrac{{15 \times 15 + 19.7 \times 19.7}}{{2 \times 9.8}} = 31.46m\]
We can find the maximum velocity by using this maximum height using the formula \[{v_{\max }} = \sqrt {{u^2} + {v^2}} = \sqrt {{{19.79}^2} + {{15}^2}} = 24.83m/s\]
The potential energy, kinetic energy, maximum height and maximum velocity is obtained as \[98J\] , \[56.25J\] , \[31.46m\] , \[24.83m/s\] respectively.

Note:
The initial velocity is not zero in this case because when the ball was thrown upwards with a force. This force causes the velocity with which the ball moves upwards resisting the force of gravity and reaches a certain height.