Question

A body projected vertically upwards with a speed v covers a distance x in the last second of its upward motion. If the projectile speed is doubled, then the distance covered in the last second will be:
(A) 2x
(B) 4x
(C) X
(D) $1\cdot 5\text{x}$

Answer 1

Hint The distance covered in the last second of upward motion is equal to the distance covered in the first second of downward direction. Using this concept, the answer can be calculated.

Complete step by step solution We know that distance covered in last second of upward journey $\left( \uparrow \right)$ is equal to distance covered in first second of downward journey $\left( \downarrow \right)$
So, in downward journey, u = 0
$\text{h}=\text{ut}+\dfrac{1}{2}\text{ g}{{\text{t}}^{2}}$
As h = x, u = 0,
So $\text{x}=\dfrac{1}{2}\text{g}{{\text{t}}^{2}}$ …… (1)
As distance x is not dependent on the velocity (as shown in equation (1) ), So even if the projectile speed is doubled, the distance covered in the first second of downward journey will remain the same.

So, the correct option is (C)

Note When the velocity is increased, the height (or distance) covered will also increase. But the distance covered in the first second of downward journey will be equal to the last second of upward journey and will be independent of velocity.