Question

A body projected vertically upwards from the top of a tower reaches the ground in t₁ second. If it is projected vertically downwards from the same point with same speed, it reaches the ground in t₂ second. If it is just dropped from the top, it reaches the ground in t second. Prove that t = $\sqrt{t_1 t_2}$.

Answer 1

t = $\sqrt{t_1 t_2}$

Answer 2

To prove the relationship $t = \sqrt{t_1 t_2}$, we analyze each scenario using the equations of motion under constant acceleration.

Let:

• $H$ be the height of the tower.
• $u$ be the initial speed of projection.
• $g$ be the acceleration due to gravity.

We choose the top of the tower as the origin and the downwards direction as positive. The displacement to reach the ground in all cases is $+H$.

Case 1: Body projected vertically upwards

Initial velocity $v_0 = -u$ (upwards is negative). Using the kinematic equation $s = v_0 t + \frac{1}{2}at^2$: $H = (-u)t_1 + \frac{1}{2}gt_1^2$ Rearranging into a quadratic equation for $t_1$: $\frac{1}{2}gt_1^2 - ut_1 - H = 0$ Solving for $t_1$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t_1 = \frac{-(-u) \pm \sqrt{(-u)^2 - 4(\frac{1}{2}g)(-H)}}{2(\frac{1}{2}g)}$ $t_1 = \frac{u \pm \sqrt{u^2 + 2gH}}{g}$ Since time $t_1$ must be positive, we take the positive root: $t_1 = \frac{u + \sqrt{u^2 + 2gH}}{g} \quad \text{(1)}$

Case 2: Body projected vertically downwards

Initial velocity $v_0 = +u$ (downwards is positive). Using the kinematic equation $s = v_0 t + \frac{1}{2}at^2$: $H = (u)t_2 + \frac{1}{2}gt_2^2$ Rearranging into a quadratic equation for $t_2$: $\frac{1}{2}gt_2^2 + ut_2 - H = 0$ Solving for $t_2$: $t_2 = \frac{-u \pm \sqrt{u^2 - 4(\frac{1}{2}g)(-H)}}{2(\frac{1}{2}g)}$ $t_2 = \frac{-u \pm \sqrt{u^2 + 2gH}}{g}$ Since time $t_2$ must be positive, we take the positive root: $t_2 = \frac{-u + \sqrt{u^2 + 2gH}}{g} \quad \text{(2)}$

Case 3: Body just dropped

Initial velocity $v_0 = 0$. Using the kinematic equation $s = v_0 t + \frac{1}{2}at^2$: $H = (0)t + \frac{1}{2}gt^2$ $H = \frac{1}{2}gt^2$ Solving for $t$: $t^2 = \frac{2H}{g} \quad \text{(3)}$

Proving the relationship

Now, multiply $t_1$ from (1) and $t_2$ from (2): $t_1 t_2 = \left(\frac{u + \sqrt{u^2 + 2gH}}{g}\right) \left(\frac{-u + \sqrt{u^2 + 2gH}}{g}\right)$ This is in the form $(A+B)(A-B) = A^2 - B^2$, where $A = \sqrt{u^2 + 2gH}$ and $B = u$. $t_1 t_2 = \frac{(\sqrt{u^2 + 2gH})^2 - u^2}{g^2}$ $t_1 t_2 = \frac{(u^2 + 2gH) - u^2}{g^2}$ $t_1 t_2 = \frac{2gH}{g^2}$ $t_1 t_2 = \frac{2H}{g}$ From equation (3), we know that $t^2 = \frac{2H}{g}$. Therefore, substituting $t^2$ into the expression for $t_1 t_2$: $t_1 t_2 = t^2$ Taking the square root of both sides (since time values are positive): $t = \sqrt{t_1 t_2}$ This proves the desired relationship.

The final answer is $\boxed{t = \sqrt{t_1 t_2}}$.

Explanation of the solution:

1. Define tower height ($H$), initial speed ($u$), and gravity ($g$).
2. Set up kinematic equations ($s = v_0 t + \frac{1}{2}at^2$) for three scenarios with origin at tower top and downwards positive:
   • Upwards projection: $H = -ut_1 + \frac{1}{2}gt_1^2 \implies t_1 = \frac{u + \sqrt{u^2 + 2gH}}{g}$
   • Downwards projection: $H = ut_2 + \frac{1}{2}gt_2^2 \implies t_2 = \frac{-u + \sqrt{u^2 + 2gH}}{g}$
   • Dropped: $H = \frac{1}{2}gt^2 \implies t^2 = \frac{2H}{g}$
3. Multiply $t_1$ and $t_2$: $t_1 t_2 = \left(\frac{u + \sqrt{u^2 + 2gH}}{g}\right) \left(\frac{-u + \sqrt{u^2 + 2gH}}{g}\right) = \frac{(u^2 + 2gH) - u^2}{g^2} = \frac{2gH}{g^2} = \frac{2H}{g}$.
4. Equate $t_1 t_2$ with $t^2$: Since $t^2 = \frac{2H}{g}$, it follows that $t_1 t_2 = t^2$.
5. Take the square root: $t = \sqrt{t_1 t_2}$.

Answer: The proof concludes that $t = \sqrt{t_1 t_2}$.