Question

A ball is thrown straight upward with a speed v from a point h meter above the ground is: taken for the ball to strike the ground is:

• A. $\frac{v}{g}\left[1+\sqrt{1+\frac{2hg}{v^2}}\right]$
• B. $\frac{v}{g}\left[1-\sqrt{1-\frac{2hg}{v^2}}\right]$
• C. $\frac{v}{g}\left[1-\sqrt{1+\frac{2hg}{v^2}}\right]$
• D. $\frac{v}{g}\left[2+\sqrt{1+\frac{2hg}{v^2}}\right]$

Answer 1

Answer: (A)

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

Answer 2

Let's set up a coordinate system with the origin at the ground and the upward direction as positive.

The initial position of the ball is $y_0 = h$.
The initial velocity of the ball is $v_0 = +v$ (since it is thrown straight upward).
The acceleration due to gravity is $a = -g$ (acting downward).

We want to find the time $t$ when the ball strikes the ground, which means its position is $y(t) = 0$.
We use the kinematic equation relating position, initial position, initial velocity, acceleration, and time:
$y(t) = y_0 + v_0 t + \frac{1}{2} a t^2$

Substitute the known values:
$0 = h + v t + \frac{1}{2} (-g) t^2$
$0 = h + vt - \frac{1}{2}gt^2$

Rearrange the equation into a standard quadratic form $At^2 + Bt + C = 0$:
$\frac{1}{2}gt^2 - vt - h = 0$

This is a quadratic equation for $t$ with coefficients $A = \frac{g}{2}$, $B = -v$, and $C = -h$.
We can solve for $t$ using the quadratic formula $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:
$t = \frac{-(-v) \pm \sqrt{(-v)^2 - 4(\frac{g}{2})(-h)}}{2(\frac{g}{2})}$
$t = \frac{v \pm \sqrt{v^2 + 2gh}}{g}$

This gives two possible solutions for $t$:
$t_1 = \frac{v + \sqrt{v^2 + 2gh}}{g}$
$t_2 = \frac{v - \sqrt{v^2 + 2gh}}{g}$

Since $v > 0$, $g > 0$, and $h > 0$, the term $\sqrt{v^2 + 2gh}$ is real and positive.
Also, $v^2 + 2gh > v^2$, which implies $\sqrt{v^2 + 2gh} > \sqrt{v^2} = v$ (assuming $v>0$).
Therefore, $v - \sqrt{v^2 + 2gh}$ is a negative quantity.
So, $t_2 = \frac{v - \sqrt{v^2 + 2gh}}{g}$ is negative. Time $t$ is the duration after the ball is thrown, so it must be positive. The negative solution represents a time before the ball was thrown, which is not relevant to the problem.

The physically relevant solution is the positive one:
$t = \frac{v + \sqrt{v^2 + 2gh}}{g}$

Now, we need to check which option matches this result. Let's rewrite the expression to match the form of option (A):
$t = \frac{v}{g} + \frac{\sqrt{v^2 + 2gh}}{g}$
We can factor out $\frac{v}{g}$ from the expression:
$t = \frac{v}{g} \left( \frac{v}{v} + \frac{\sqrt{v^2 + 2gh}}{v} \right)$
$t = \frac{v}{g} \left( 1 + \frac{\sqrt{v^2 + 2gh}}{\sqrt{v^2}} \right)$ (assuming $v>0$)
$t = \frac{v}{g} \left( 1 + \sqrt{\frac{v^2 + 2gh}{v^2}} \right)$
$t = \frac{v}{g} \left( 1 + \sqrt{\frac{v^2}{v^2} + \frac{2gh}{v^2}} \right)$
$t = \frac{v}{g} \left[ 1 + \sqrt{1 + \frac{2gh}{v^2}} \right]$

This expression matches option (A).