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. $$5.84m{s}^{-1}$$
B. $$49m{s}^{-1}$$
C. $$\sqrt{98}m{s}^{-1}$$
D. $$98m{s}^{-1}$$

Answer 1

Hint: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).
$$0=u-gt$$
$$\Rightarrow u=gt$$
$$\Rightarrow u=9.8\times 5$$
$$\therefore u=49m{s}^{-1}$$

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 $$s=u+{\displaystyle \frac{1}{2}}\left(2n-1\right)$$. 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.