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AP EAMCET · PHYSICS · Motion In Two Dimensions

A body is projected horizontally from the top of a tall tower with a velocity of \(30 \mathrm{~ms}^{-1}\). At time \(t_1\), its horizontal and vertical components of the velocity are equal and at time \(t_2\), its horizontal and vertical displacements are equal. Then \(t_2-t_1\) is (take, \(g=10 \mathrm{~ms}^{-2}\) )

  1. A \(1 \mathrm{~s}\)
  2. B \(1.5 \mathrm{~s}\)
  3. C \(2 \mathrm{~s}\)
  4. D \(3 \mathrm{~s}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3 \mathrm{~s}\)

Step-by-step Solution

Detailed explanation

As per first condition, \[ \begin{gathered} 30=10 t_1 \\ t_1=3 \mathrm{~s} \end{gathered} \] As per second condition, \[ \begin{aligned} 30 t_2 & =\frac{1}{2} \times 10 t_2^2=t_2=6 \mathrm{~s} \\ \therefore \quad t_2-t_1 & =3 \mathrm{~s} \end{aligned} \]
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