Question

A ball is projected vertically upwards from ground. It reaches a height 'h' in time $t_1$, continues its motion and then takes a time $t_2$ to reach ground. The height h in terms of g, $t_1$ and $t_2$ is (g = acceleration due to gravity)

• A. $\frac{1}{2}g\,t_1t_2$

Answer 1

Answer: (A)

$\frac{1}{2} g\,t_1\,t_2$

Answer 2

Solution:

Let the time when the ball passes height $h$ (on the upward journey) be $t_1$, so its displacement is

$$h = u\,t_1 - \frac{1}{2}g\,t_1^2 \quad \text{(1)}$$

After passing $h$, the ball continues its motion and hits the ground at time $t_1+t_2$. At that instant, the displacement from the ground is zero:

$$0 = u\,(t_1+t_2) - \frac{1}{2}g\,(t_1+t_2)^2 \quad \text{(2)}$$

From (2), solving for $u$:

$$u\,(t_1+t_2)= \frac{1}{2}g\,(t_1+t_2)^2 \quad\Longrightarrow\quad u= \frac{1}{2}g\,(t_1+t_2)$$

Substitute $u$ into (1):

$$h = \frac{1}{2}g\,(t_1+t_2)t_1 - \frac{1}{2}g\,t_1^2$$ $$\implies h = \frac{1}{2}g \left[t_1^2+t_1t_2 -t_1^2\right] = \frac{1}{2}g\,t_1t_2$$

Core Explanation:

1. Use displacement at time $t_1$: $h = ut_1 - \frac{1}{2}gt_1^2$.

2. At ground (time $t_1+t_2$): $0 = u(t_1+t_2) - \frac{1}{2}g(t_1+t_2)^2$ to get $u = \frac{1}{2}g(t_1+t_2)$.

3. Substitute $u$ into the first equation to get $h = \frac{1}{2}gt_1t_2$.