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

Step 1: Use the distance-in-nth-second formula.
Under constant acceleration, the distance covered in the $n$th second is $S_n=u+\dfrac{a}{2}(2n-1)$. Here $u=60$ m/s upward and $a=-g=-10$ m/s$^2$.
Step 2: First second ($n=1$).
$S_1=60+\dfrac{-10}{2}(2\cdot1-1)=60-5=55$ m.
Step 3: Second second ($n=2$).
$S_2=60+\dfrac{-10}{2}(2\cdot2-1)=60-15=45$ m.
Step 4: Third second ($n=3$).
$S_3=60+\dfrac{-10}{2}(2\cdot3-1)=60-25=35$ m.
Step 5: Form the ratio.
$S_1:S_2:S_3=55:45:35$.
Step 6: Reduce.
Dividing each by $5$ gives $11:9:7$.
\[ \boxed{11:9:7} \]