Question

A ball is projected vertically up with speed \( V_0 \) from a certain height H. When the ball reaches the ground the speed is \( 3V_0 \). The time taken by the ball to reach the ground and height H respectively are (g = acceleration due to gravity)

Hint:

For vertical motion, remember that the final velocity of an object thrown upwards will be greater than its initial velocity when falling back down, due to the acceleration due to gravity.

Answer 1

We are given that the ball is projected vertically upward with speed \( V_0 \) from a height \( H \), and when it reaches the ground, its speed is \( 3V_0 \). We need to find the time taken to reach the ground and the value of height \( H \). Using the equations of motion:
1. \( v^2 = u^2 + 2gh \), where:
- \( v = 3V_0 \) (final velocity)
- \( u = V_0 \) (initial velocity)
- \( h = H \) (height traveled)
So, \( (3V_0)^2 = (V_0)^2 + 2gH \), which simplifies to:
\[ 9V_0^2 = V_0^2 + 2gH \] \[ 8V_0^2 = 2gH \] \[ H = \frac{4V_0^2}{g} \] 2. To find the time \( t \), we use the formula:
\[ v = u + gt \] \[ 3V_0 = V_0 + gt \] Solving for \( t \), we get: \[ 2V_0 = gt \] \[ t = \frac{2V_0}{g} \]