Question

With what speed can a body be thrown upwards so that the distance traversed in ${5^{th}}$ second and ${6^{th}}$ seconds are equal?
A. $5.84\,m{s^{ - 1}}$
B. $49\,m{s^{ - 1}}$
C. $\sqrt {98}\,m{s^{ - 1}}$
D. $98\,m{s^{ - 1}}$

Answer 1

in this question the condition that distance traversed in ${5^{th}}$ second and ${6^{th}}$ seconds are equal only conceivable if the body decelerates for 4 to 5 seconds and then accelerates for 5 to 6 seconds. This state is suitable for projectile motion if the body moves upward in 4 to 5 seconds and downward in 5 to 6 seconds. After understanding this concept in detail we will apply the first equation of motion and come to an answer.

Formula used:
$v = u + at$
Where $v$- final velocity, $u$- initial velocity, $a$- acceleration and $t$-time.

Complete step by step answer:
Initially the ball is thrown with 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 highest point, and it reaches there at ${5^{th}}$ second.Here time of flight is equal to time of descent. Hence the body is at its highest point at ${5^{th}}$ second.

The body when thrown up slowly decelerates when it moves upwards and becomes 0 at the top most point. It again accelerates from that point and falls downwards. The same condition has to be applied here. From ${4^{th}}$ to ${5^{th}}$ second the body decelerates at ${5^{th}}$ second its speed becomes ‘0’ and again from ${5^{th}}$ second to ${6^{th}}$ second the body accelerates.

Hence the time required by the body upwards (during ${4^{th}}$ to ${5^{th}}$ second) is equal to the time required by it to move downwards (from ${5^{th}}$ second to ${6^{th}}$ second),which has to be 5seconds.Look at the diagram to get a better idea:

Hence we will apply first equation of motion:
$v = u + at$
We know that at the highest point of a projectile the velocity is zero. Hence
$0 = u + at$
Also a will be equal to acceleration due to gravity ‘g’. (it will be –g because the body is thrown upwards).
$0 = u - gt$
$\Rightarrow u = gt$
$\Rightarrow u = 9.8 \times 5$
$\therefore u = 49\,m{s^{ - 1}}$

Hence the correct answer is option B.

Note: Students make a very common mistake in solving distance covered in ${n^{th}}$second problems, they tend to apply the formula $s = u + \dfrac{1}{2}\left( {2n - 1} \right)$. This formula is correct but this definitely cannot be used to solve this problem, because if $u$ equates distances. All variables get canceled out. Also acceleration for a body moving upwards has to be negative.