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:

• A. \( \frac{V_0}{g} \), \( \frac{V_0^2}{2g} \)
• B. \( \frac{V_0}{g} \), \( \frac{3V_0^2}{2g} \)
• C. \( \frac{2V_0}{g} \), \( \frac{V_0^2}{2g} \)
• D. \( \frac{3V_0}{g} \), \( \frac{3V_0^2}{2g} \)

Hint:

For vertical motion under gravity, use the equations of motion to relate velocity, acceleration, and displacement. Remember to carefully consider the signs of quantities like acceleration when working with upward and downward motion.

Answer 1

The Correct Option is B

Let the initial velocity be \( V_0 \) and the height be \( H \). The ball is projected upwards, and it reaches the ground with a speed of \( 3V_0 \). We can use the equation of motion for vertical displacement and velocity. The equation relating velocity, acceleration, and displacement is: \[ v^2 = u^2 + 2a s \] where: - \( v \) is the final velocity, - \( u \) is the initial velocity, - \( a \) is the acceleration (which is \( -g \), the acceleration due to gravity, as the ball is moving upwards), - \( s \) is the displacement (which in this case is \( H \)). At the point when the ball reaches the ground, we have: - Final velocity \( v = 3V_0 \), - Initial velocity \( u = V_0 \), - Displacement \( s = H \), - Acceleration \( a = -g \). Using the equation: \[ (3V_0)^2 = (V_0)^2 + 2(-g)H \] Simplifying: \[ 9V_0^2 = V_0^2 - 2gH \] \[ 9V_0^2 - V_0^2 = 2gH \] \[ 8V_0^2 = 2gH \] \[ H = \frac{4V_0^2}{g} \] Thus, the height \( H \) is \( \frac{4V_0^2}{g} \). Now, we need to find the time taken by the ball to reach the ground. The time \( t \) can be found from the equation: \[ v = u + at \] Using the same values for \( v = 3V_0 \), \( u = V_0 \), and \( a = -g \), we get: \[ 3V_0 = V_0 - g t \] Solving for \( t \): \[ 3V_0 - V_0 = g t \] \[ 2V_0 = g t \] \[ t = \frac{2V_0}{g} \] Thus, the time taken by the ball to reach the ground is \( \frac{2V_0}{g} \). So, the correct answer is (B) \( \frac{V_0}{g} \), \( \frac{3V_0^2}{2g} \).