Question

A body is thrown vertically upwards from earth with velocity \(60\text{ ms}^{-1}\). The ratio of displacements during first, second and third seconds is (\(g=10\text{ ms}^{-2}\))

• A. \(1:3:5\)
• B. \(11:9:7\)
• C. \(1:1:1\)
• D. \(1:2:3\)

Hint:

Nth second formula: \[ S_n=u+\frac{a}{2}(2n-1) \] For upward motion take acceleration negative.

Answer 1

The Correct Option is B

Concept: Distance covered in nth second under uniform acceleration: \[ S_n=u+\frac{a}{2}(2n-1) \] For upward motion acceleration is negative. \[ a=-g=-10 \]

Step 1: Distance in first second.
\[ S_1=60+\frac{-10}{2}(1) \] \[ S_1=60-5=55 \]

Step 2: Distance in second second.
\[ S_2=60+\frac{-10}{2}(3) \] \[ S_2=60-15=45 \]

Step 3: Distance in third second.
\[ S_3=60+\frac{-10}{2}(5) \] \[ S_3=60-25=35 \] Thus ratio becomes \[ 55:45:35 \] Dividing by 5 \[ 11:9:7 \] Hence \[ \boxed{11:9:7} \]