Question

A ball is projected upwards from a height h above the surface of the earth with velocity v. The time at which the ball strikes the ground is

• A. $\frac{v}{g} + \sqrt{\frac{2h}{g}}$
• B. $\frac{2v}{g} - \sqrt{\frac{2h}{g}}$
• C. $\frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^2}} \right]$
• D. None of these

Answer 1

Answer: (C)

$\frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^2}} \right]$

Answer 2

The problem describes a ball projected upwards from a height 'h' with an initial velocity 'v'. We need to find the total time until the ball strikes the ground. We can use the equations of motion under constant acceleration due to gravity.

Let's set up a coordinate system:

• Let the initial position (height 'h' above the ground) be the origin (y=0).
• Let the upward direction be positive.

Given:

• Initial velocity, $u = +v$ (upwards)
• Acceleration due to gravity, $a = -g$ (downwards)
• Displacement when the ball strikes the ground, $s = -h$ (from the initial height to the ground, which is 'h' units below the starting point).
• Time taken, $t = ?$

Using the kinematic equation:

$s = ut + \frac{1}{2}at^2$

Substitute the values:

$-h = vt + \frac{1}{2}(-g)t^2$

$-h = vt - \frac{1}{2}gt^2$

Rearrange the equation into a standard quadratic form $At^2 + Bt + C = 0$:

$\frac{1}{2}gt^2 - vt - h = 0$

Multiply the entire equation by 2 to clear the fraction:

$gt^2 - 2vt - 2h = 0$

This is a quadratic equation for 't'. We can solve for 't' using the quadratic formula:

$t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Here, $A = g$, $B = -2v$, and $C = -2h$.

Substitute these into the formula:

$t = \frac{-(-2v) \pm \sqrt{(-2v)^2 - 4(g)(-2h)}}{2(g)}$

$t = \frac{2v \pm \sqrt{4v^2 + 8gh}}{2g}$

Factor out 4 from the term under the square root:

$t = \frac{2v \pm \sqrt{4(v^2 + 2gh)}}{2g}$

$t = \frac{2v \pm 2\sqrt{v^2 + 2gh}}{2g}$

Divide all terms in the numerator and denominator by 2:

$t = \frac{v \pm \sqrt{v^2 + 2gh}}{g}$

Since time 't' must be a positive value, we choose the positive root. The term $\sqrt{v^2 + 2gh}$ is always greater than 'v' (as $2gh > 0$), so $v - \sqrt{v^2 + 2gh}$ would be negative. Therefore, we take the positive sign:

$t = \frac{v + \sqrt{v^2 + 2gh}}{g}$

To match the given options, we can factor out $\frac{v}{g}$ from the expression. To do this, we need to factor out $v^2$ from inside the square root:

$t = \frac{v}{g} \left( 1 + \frac{\sqrt{v^2 + 2gh}}{v} \right)$

$t = \frac{v}{g} \left( 1 + \sqrt{\frac{v^2 + 2gh}{v^2}} \right)$

$t = \frac{v}{g} \left( 1 + \sqrt{\frac{v^2}{v^2} + \frac{2gh}{v^2}} \right)$

$t = \frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^2}} \right]$

This expression matches option (3).