Question

A ball is thrown vertically upwards. It was observed at a height $h$ twice with a time interval $\Delta t$. The initial velocity of the ball is

• A. $\sqrt{8gh+g^2(\Delta t)^2}$
• B. $\sqrt{8gh+(\frac{g\Delta t}{2})^2}$
• C. $\frac{1}{2}\sqrt{8gh+g^2(\Delta t)^2}$
• D. $\sqrt{8gh+4g^2(\Delta t)^2}$

Answer 1

Answer: (C)

(C) $\frac{1}{2}\sqrt{8gh+g^2(\Delta t)^2}$

Answer 2

The motion of a ball thrown vertically upwards is governed by the equation of motion under constant acceleration due to gravity. Let the initial velocity of the ball be $u$, the height be $h$, and the acceleration due to gravity be $g$. We take the upward direction as positive, so the acceleration is $-g$.

The equation relating displacement ($h$), initial velocity ($u$), time ($t$), and acceleration ($-g$) is:

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

Rearrange this equation into a standard quadratic form in terms of time $t$:

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

Multiply by 2 to clear the fraction:

$gt^2 - 2ut + 2h = 0$

Let $t_1$ and $t_2$ be the two instants of time when the ball is at height $h$. These are the two roots of the quadratic equation. According to Vieta's formulas for a quadratic equation $at^2 + bt + c = 0$, the sum of the roots ($t_1 + t_2$) and the product of the roots ($t_1 t_2$) are given by:

Sum of roots: $t_1 + t_2 = -\frac{(-2u)}{g} = \frac{2u}{g}$

Product of roots: $t_1 t_2 = \frac{2h}{g}$

We are given that the time interval between the two observations at height $h$ is $\Delta t$. Assuming $t_2 > t_1$, we have:

$t_2 - t_1 = \Delta t$

Now we use the algebraic identity that relates the sum, difference, and product of two numbers:

$(t_2 + t_1)^2 - (t_2 - t_1)^2 = 4t_1 t_2$

Substitute the expressions for the sum, difference, and product of $t_1$ and $t_2$ into this identity:

$(\frac{2u}{g})^2 - (\Delta t)^2 = 4(\frac{2h}{g})$

$\frac{4u^2}{g^2} - (\Delta t)^2 = \frac{8h}{g}$

Now, we need to solve for $u$:

$\frac{4u^2}{g^2} = \frac{8h}{g} + (\Delta t)^2$

To isolate $u^2$, multiply both sides by $g^2$:

$4u^2 = g^2(\frac{8h}{g} + (\Delta t)^2)$

$4u^2 = 8hg + g^2(\Delta t)^2$

Divide by 4:

$u^2 = \frac{8hg + g^2(\Delta t)^2}{4}$

$u^2 = \frac{1}{4}(8hg + g^2(\Delta t)^2)$

Take the square root of both sides to find $u$:

$u = \sqrt{\frac{1}{4}(8hg + g^2(\Delta t)^2)}$

$u = \frac{1}{2}\sqrt{8hg + g^2(\Delta t)^2}$

Comparing this result with the given options, it matches option (C).