Question

the vertical height of P above the ground is twice that of Q. A particle is projected downward with a speed of 9.8m per second from P and simultaneously another particle is projected with same velocity from Q. both particle reachthe ground simultaneously.the time taken to reach the ground is-

Answer 1

the vertical height of $\mathsf{P}$ above the ground is twice that of Q. A particle is projected downward with a speed of 9.8m per second from P and simultaneously another particle is projected with same velocity from Q. both particle reach the ground simultaneously. the time taken to reach the ground is-

Solution:

$V_{Q} = V_{P} = V = 9.8\frac{m}{s}$ - speed of the particles

$t_{Q} = t_{P} = t -$ time taken to reach the ground

Equation of the motion for particle at the height $P$ above the ground on Y-axis:

$$P = V t + \frac {g t ^ {2}}{2}$$

Equation of the motion for particle at the height $Q$ above the ground on Y-axis (velocity is directed upwards):

$$Q = - V t + \frac {g t ^ {2}}{2}$$ $$P = 2 \cdot Q$$

(1) and (2) in (3):

$$2 \left(- V t + \frac {g t ^ {2}}{2}\right) = V t + \frac {g t ^ {2}}{2}$$ $$- 2 V t + g t ^ {2} = V t + \frac {g t ^ {2}}{2}$$ $$3 V = \frac {g t}{2}$$ $$t = \frac {6 V}{g} = \frac {6 \cdot 9 . 8 \frac {m}{s}}{9 . 8 \frac {m}{s ^ {2}}} = 6 s$$

Answer: time taken to reach the ground is 6s.