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KCET · Physics · Motion In One Dimension

In a lift moving up with an acceleration of \(5 \mathrm{~ms}^{-2}\), a ball is dropped from a height of \(1.25 \mathrm{~m}\). The time taken by the ball to reach the floor of the lift is ... (nearly) \(\left(g=10 \mathrm{~ms}^{-2}\right)\)

  1. A \(0.3 \mathrm{~s}\)
  2. B \(0.2 \mathrm{~s}\)
  3. C \(0.16 \mathrm{~s}\)
  4. D \(0.4 \mathrm{~s}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(0.4 \mathrm{~s}\)

Step-by-step Solution

Detailed explanation

Distance travelled by ball during fall \(=1.25 \mathrm{~m}\)
By using relation \(s=u t+\frac{1}{2} a t^{2}\)
Here, \(u=0, \quad s=1.25 \mathrm{~m}\),
\(a =(g+a)=(10+5)=15 \mathrm{~m} / \mathrm{s}^{2} \)
\(\Rightarrow 1.25 =\frac{1}{2} \times 15 \times t^{2}, \Rightarrow t^{2}=\frac{2 \times 1.25}{15} \)
\(t^{2} =0.16 \quad \Rightarrow t=0.4 \mathrm{~s}\)