Question

With what speed should a body be thrown upwards so that the distances covered in the \(5^{th}\) second and \(6^{th}\) second are equal?

• A. \(75\ \text{m/s}\)
• B. \(\sqrt{98}\ \text{m/s}\)
• C. \(49\ \text{m/s}\)
• D. \(19.8\ \text{m/s}\)

Hint:

For upward motion, velocity at the highest point is zero. Use \(v=u-gt\) to find initial speed.

Answer 1

The Correct Option is C

Concept: For vertical upward motion, the body slows down due to gravity. Equal distances in two consecutive seconds around the highest point occur when the highest point is reached between those seconds.

Step 1: The body is thrown upward with speed \(u\).
At the highest point, velocity becomes zero. Using: \[ v=u-gt \] At the highest point: \[ v=0 \] So: \[ 0=u-gt \] \[ u=gt \]

Step 2: Since distances in the \(5^{th}\) and \(6^{th}\) seconds are equal, the highest point lies at the boundary between the \(5^{th}\) and \(6^{th}\) second. Therefore: \[ t=5\ \text{s} \]

Step 3: Use \(g=9.8\ \text{m/s}^2\). \[ u=gt \] \[ u=9.8\times 5 \] \[ u=49\ \text{m/s} \] Therefore, \[ \boxed{49\ \text{m/s}} \]