Question: A particle $P_1$ is projected vertically upwards, from horizontal ground, with a speed of $30$ m s$^{-1}$. At the same instant another particle $P_2$ is projected vertically upwards from the top of a tower of height $25$ m, with a speed of $10$ m s$^{-1}$. Find
(i) the time for which $P_1$ is higher than the top of the tower,
(ii) the velocities of the particles at the instant when the particles are at the same height,
(iii) the time for which $P_1$ is higher than $P_2$ and is moving upwards.
1. **Problem Statement:**
We have two particles projected vertically upwards:
- Particle $P_1$ from ground with initial velocity $u_1 = 30$ m/s.
- Particle $P_2$ from top of tower of height $h = 25$ m with initial velocity $u_2 = 10$ m/s.
We use acceleration due to gravity $g = 9.8$ m/s$^2$ acting downwards.
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2. **Formulas and Rules:**
Height at time $t$ for vertical motion under gravity:
$$ s = ut - \frac{1}{2}gt^2 $$
Velocity at time $t$:
$$ v = u - gt $$
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3. **(i) Time for which $P_1$ is higher than the top of the tower:**
We want $s_1 > 25$ m where
$$ s_1 = 30t - \frac{1}{2} \times 9.8 \times t^2 = 30t - 4.9t^2 $$
Set $s_1 = 25$ to find boundary times:
$$ 30t - 4.9t^2 = 25 $$
Rearranged:
$$ 4.9t^2 - 30t + 25 = 0 $$
Use quadratic formula:
$$ t = \frac{30 \pm \sqrt{30^2 - 4 \times 4.9 \times 25}}{2 \times 4.9} = \frac{30 \pm \sqrt{900 - 490}}{9.8} = \frac{30 \pm \sqrt{410}}{9.8} $$
Calculate:
$$ \sqrt{410} \approx 20.25 $$
So,
$$ t_1 = \frac{30 - 20.25}{9.8} = \frac{9.75}{9.8} \approx 0.995 \text{ s} $$
$$ t_2 = \frac{30 + 20.25}{9.8} = \frac{50.25}{9.8} \approx 5.13 \text{ s} $$
$P_1$ is above $25$ m between $t_1$ and $t_2$.
**Answer:** $P_1$ is higher than the tower for approximately $0.995$ s to $5.13$ s.
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4. **(ii) Velocities of particles when they are at the same height:**
Let $t$ be the time when $P_1$ and $P_2$ are at the same height.
Heights:
$$ s_1 = 30t - 4.9t^2 $$
$$ s_2 = 25 + 10t - 4.9t^2 $$
Set equal:
$$ 30t - 4.9t^2 = 25 + 10t - 4.9t^2 $$
Cancel $-4.9t^2$ on both sides:
$$ 30t = 25 + 10t $$
$$ 30t - 10t = 25 $$
$$ 20t = 25 $$
$$ t = \frac{25}{20} = 1.25 \text{ s} $$
Velocities at $t=1.25$ s:
$$ v_1 = 30 - 9.8 \times 1.25 = 30 - 12.25 = 17.75 \text{ m/s upwards} $$
$$ v_2 = 10 - 9.8 \times 1.25 = 10 - 12.25 = -2.25 \text{ m/s (downwards)} $$
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5. **(iii) Time for which $P_1$ is higher than $P_2$ and moving upwards:**
$P_1$ is moving upwards when $v_1 > 0$:
$$ 30 - 9.8t > 0 \Rightarrow t < \frac{30}{9.8} \approx 3.06 \text{ s} $$
Find when $P_1$ is higher than $P_2$:
Set $s_1 > s_2$:
$$ 30t - 4.9t^2 > 25 + 10t - 4.9t^2 $$
Cancel $-4.9t^2$:
$$ 30t > 25 + 10t $$
$$ 20t > 25 $$
$$ t > 1.25 \text{ s} $$
Combine both conditions:
$$ 1.25 < t < 3.06 $$
**Answer:** $P_1$ is higher than $P_2$ and moving upwards between $1.25$ s and $3.06$ s.