Question

A cricket ball and a tennis ball are simultaneously thrown upwards from the ground with equal initial speeds. Ignoring the air resistance, which of the following statements is correct?
A. Both the balls travel the same distance at the same time.
B. Tennis ball rises the highest.
C. Tennis ball is in flight the longest.
D. Tennis ball has a smaller impact speed.

Answer 1

Hint: We will make use of the formula of the law of motion to find the distance covered by the balls. The weight of the balls doesn’t play any role in the absence of air resistance. Thus, by considering the gravitational acceleration and the same initial velocity, we will prove that the distance covered by both the balls are the same.

Formula used: \[{{v}^{2}}-{{u}^{2}}=2as\]

Complete step by step answer:
The answer to this question completely depends on the initial speed of the objects and their weight.
As, we are given the same initial speed of a cricket ball and a tennis ball, so, we can derive the equations as follows.
The final velocity of both the balls will be zero, as they will hit the ground at the end of their motion.
The law of motion equation is given as follows.
\[{{v}^{2}}-{{u}^{2}}=2as\]
Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance covered.
Consider the notation 1 for the cricket ball and the notation 2 for the tennis ball. As these balls are falling, thus instead of the normal acceleration, the acceleration due to gravity acts on these.
So, we have,
For cricket ball, \[{{v}^{2}}_{1}-{{u}^{2}}_{1}=2{{g}_{1}}{{s}_{1}}\]
For tennis ball, \[{{v}^{2}}_{2}-{{u}^{2}}_{2}=2{{g}_{2}}{{s}_{2}}\]
Further compute the equations
For cricket ball,
\[\begin{align}
  & 0-{{u}^{2}}_{1}=2g{{s}_{1}} \\
 & \Rightarrow {{s}_{1}}=-\dfrac{{{u}_{1}}^{2}}{2g} \\
\end{align}\]
For tennis ball,
\[\begin{align}
  & 0-{{u}^{2}}_{2}=2g{{s}_{2}} \\
 & \Rightarrow {{s}_{2}}=-\dfrac{{{u}_{2}}^{2}}{2g} \\
\end{align}\]
The negative sign indicates the direction. As the initial velocities are the same, less us consider it as ‘u’. \[{{u}_{1}}={{u}_{2}}=u\]
So, we get,
\[\begin{align}
  & {{s}_{1}}=-\dfrac{{{u}^{2}}}{2g},{{s}_{2}}=-\dfrac{{{u}^{2}}}{2g} \\
 & \Rightarrow {{s}_{1}}={{s}_{2}} \\
\end{align}\]

Therefore, the distance covered by both the balls are the same. So, the correct answer is “Option A”

Additional Information: Consider the presence of air resistance. The more the weight of a body, the more will be the amount of air resistance acting on that body. In turn, the faster the speed with which the body reaches the ground. Thus, the body with more weight will reach the ground faster when compared to the body with the lightweight.

Note: The main point to remember in this case is the ‘air resistance’. The amount of the air resistance experienced by the falling body depends on the speed with which the body is falling, its shape, area of cross-section/weight and the air density. Here the air density is considered, as it varies with the height (altitude), the temperature and the humidity (water vapour of the atmosphere).