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

The ball is thrown straight upward from a point $h$ meters above the ground with an initial speed $v$. We need to find the time taken for the ball to strike the ground.

Let's set up a coordinate system with the origin at the ground level and the positive y-axis pointing upward.

The initial position of the ball is $y_0 = h$.

The initial velocity of the ball is $v_0 = +v$ (since it is thrown upward).

The acceleration due to gravity acts downward, so $a = -g$.

The final position of the ball when it strikes the ground is $y = 0$.

We use the kinematic equation that relates position, initial position, initial velocity, acceleration, and time: $y = y_0 + v_0 t + \frac{1}{2} a t^2$

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

Rearrange the equation to form a quadratic equation in terms of $t$: $\frac{1}{2}gt^2 - vt - h = 0$ Multiply by 2 to clear the fraction: $gt^2 - 2vt - 2h = 0$

This is a quadratic equation of the form $At^2 + Bt + C = 0$, where $A=g$, $B=-2v$, and $C=-2h$. We can solve for $t$ using the quadratic formula: $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

Substitute the values of $A$, $B$, and $C$: $t = \frac{-(-2v) \pm \sqrt{(-2v)^2 - 4(g)(-2h)}}{2(g)}$ $t = \frac{2v \pm \sqrt{4v^2 + 8gh}}{2g}$

We can factor out 4 from the term inside the square root: $t = \frac{2v \pm \sqrt{4(v^2 + 2gh)}}{2g}$ $t = \frac{2v \pm 2\sqrt{v^2 + 2gh}}{2g}$

Now, divide the numerator and the denominator by 2: $t = \frac{v \pm \sqrt{v^2 + 2gh}}{g}$

We have two possible solutions for $t$. Since time $t$ must be positive, we need to evaluate the two roots. The term $\sqrt{v^2 + 2gh}$ is positive. Since $v > 0$, $g > 0$, and $h > 0$, we have $v^2 + 2gh > v^2$. Taking the square root, we get $\sqrt{v^2 + 2gh} > \sqrt{v^2} = v$. So, $v - \sqrt{v^2 + 2gh}$ will be negative. And $v + \sqrt{v^2 + 2gh}$ will be positive.

Since the time taken must be positive, we take the positive root: $t = \frac{v + \sqrt{v^2 + 2gh}}{g}$

Now, we need to express this result in the form given in the options. The options have a factor of $\frac{v}{g}$ outside a bracket. Let's factor out $\frac{v}{g}$ from our expression for $t$: $t = \frac{v}{g} \left( \frac{v + \sqrt{v^2 + 2gh}}{v} \right)$ $t = \frac{v}{g} \left( \frac{v}{v} + \frac{\sqrt{v^2 + 2gh}}{v} \right)$ $t = \frac{v}{g} \left( 1 + \sqrt{\frac{v^2 + 2gh}{v^2}} \right)$ $t = \frac{v}{g} \left( 1 + \sqrt{\frac{v^2}{v^2} + \frac{2gh}{v^2}} \right)$ $t = \frac{v}{g} \left( 1 + \sqrt{1 + \frac{2gh}{v^2}} \right)$

This expression matches option (A).