Question

The distance travelled by a body during the last second of its upward journey is d, when the body is projected with certain velocity vertically up. If the velocity of projection is doubled, the distance travelled by the body during the last second of its upward journey is:
A) 2d
B) 4d
C) $\dfrac{d}{2}$
D) d

Answer 1

Equation of motion has the solution of this question. S=ut+$\dfrac{1}{2}a{t^2}$ where, S is the distance and u is the initial speed of an object, a is the acceleration and t is time. But here the body is travelling in projectile motion which means it will suffer acceleration due to gravity. By the eqn of last second and just before of it we will get the distance travelled by it.

Step by step solution:
Step 1:
Equation o f motion: These equations are used to derive the components like displacement(s), velocity (initial and final), time (t) and acceleration (a). Therefore they can only be applied when acceleration is constant and motion is a straight line.
These equations are,
v = u + at.
${v^2}$ =${u^2}$ +2as
S=ut+$\dfrac{1}{2}a{t^2}$
Where, S is the distance and u is the initial speed of an object, v is the final speed of an object, and a is the acceleration and t is time.
Step2:
We are given that the distance travelled by a body during the last second of its upward journey is d, when the body is projected with certain velocity vertically up and we need to find distance when velocity is doubled up.
Now, Let the speed of projection be u and the time taken to reach maximum height be t.
${S_t}$ =ut−$\dfrac{1}{2}g{t^2}$……(1) here, negative sign because gravity is acting on the body.
Distance just before the last second before attaining the maximum height is :
${S_{t - 1}}$ =u(t-1)−$\dfrac{1}{2}g{\left( {t - 1} \right)^2}$…….(2)
Now, subtracting (1) from (2) we will get the distance covered by a body in normal conditions.
Subtracting (1) from (2) we get:
${S_t}$−${S_{t - 1}}$=u−gt+$\dfrac{g}{2}$ …..(3)
For a vertically projected body, u=gt (because ’t’ is the time when it reaches max. height), hence
${S_t}$−${S_{t - 1}}$=$\dfrac{g}{2}$ …….(4)
Hence from eqn. (4) we can say that this shows the distance traveled by a body projected vertically up is independent of the initial speed. So, even if the body is projected with double the speed, the body covers the same distance, in the last second.

Option D is correct.

Note:
Point of mistake:
At the highest point remember that the acceleration never becomes zero and horizontal velocity is constant. When the projectile hits the target the final velocity is not equal to zero.
In equation (1) it is given that there is a negative sign because of the gravity. As we know the tendency of gravity to attract so it will also want to bring the body to rest and attract toward it and that will be in the opposite direction.