Question

A body $A$ is dropped from a height $h$ and the other body $B$ is thrown horizontally with a velocity of $10m{s}^{-1}$ from the same height simultaneously. If the time taken by the body to reach the ground is $3s$, arrange the following steps in a proper sequence to find the velocity of the body $B$, in vertical direction when it is at a vertical height of $20m$ above the ground.

a) Find the displacement covered by the body $B$ when it is at a height of $20m$.

b) Calculate the vertical height from the ground from which both the bodies are falling from $t=0$

c) Write the required equation of motion and calculate the velocity at the given point.

d) Analyse the relation between the motions of two bodies.

• A. d c a b
• B. d b a c
• C. a b c d
• D. d b c a

Answer 1

Answer: (B)

d b a c

(i) The vertical motion of a horizontal projectile is same as that of a freely falling body.
$\therefore$ The time taken to reach the ground for the horizontal projectile $t=3s$.
(ii) $\therefore$ The vertical height from the ground $h=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 3^2=45m$
$h=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 3^2=45m$
(iii) The displacement covered by B when it is at a height of $20m$ from the ground,
$h_1=45-20=25m$
(iv) $\therefore V^2=u^2+2g{h}_1$
$V^2=2\times 10\times 25=500$
$V=\sqrt{500}=10\sqrt{5}m{s}^{-1}$

Therefore the order is d b a c.