Question

With what speed should a body be thrown upwards so that the distance traversed in the 5^th second and 6^th second are equal?

• A. 5.84 m/s
• B. 49 m/s
• C. $\sqrt{98}$ m/s
• D. 98 m/s

Answer 1

The correct option is B

49 m/s

Step 1: Given parameters

Time periods = 5 sec ad 6 sec

The acceleration due to gravity, $g=9.8m/s^2$

Step 2: Formula used

Use the first equation of motion given as

$v=u-gt$

Where, $v$ is final velocity, $u$ is the initial velocity, $g$ is the acceleration due to gravity and $t$ is the time period.

Here, the acceleration due to gravity is negative as the body is moving up.

Step 3: Calculate the initial velocity

Consider the diagram illustrated below.

At 5^th sec and 6th-sec, distance traveled = $s$ and in 4 sec to 5-sec, distance traveled = $s$

Initially, the ball is thrown at a certain speed. We are supposed to find this speed.
The condition that distance traversed in 5^th second and 6^th seconds are equal only when the body is at the highest point, and it reaches there at the 5^th second.

Here, the time of flight is equal to the time of descent. Hence the body is at its highest point at the 5^th second.

This is only possible if the body decelerated in 4 seconds to 5 seconds and accelerated in 5 seconds to 6 seconds.

This condition is suited to projectile motion

In 4 sec to 5-sec body go upward and in 5 sec to 6-sec body go downward

Hence, at 5-sec body is at the top and its vertical velocity is zero.

Use the first equation of motion to calculate the initial velocity,

$$\begin{gathered} v=u-gt \\ 0=u-gt \\ u=gt \\ u=9.81\times 5 \\ u=49.01m/s \end{gathered}$$

Hence, the initial velocity is 49 m/s.

Therefore, option (B) is correct.