Question

Two bodies are thrown vertically upward with the same initial velocity of 98m/s but 4 second apart.How long after the first one is thrown will they meet?

Answer 1

Two bodies are thrown vertically upward with the same initial velocity of $98\mathrm{m / s}$ but 4 second apart. How long after the first one is thrown will they meet?

Solution:

t - time passed after the first body was thrown;

$t_m$ - time before the meeting of the bodies;

$\Delta t = 4s - \text{delay time;}$

$V = 98\frac{m}{s} - \text{velocity of the bodies.}$

Equation of motion for the first body:

$$y _ {1} = V t - \frac {g t ^ {2}}{2}$$

Equation of motion for the second body:

$$y _ {2} = V (t - \Delta t) - \frac {g (t - \Delta t) ^ {2}}{2}$$

The condition of the meeting:

$$y _ {1} \left(t _ {m}\right) = y _ {2} \left(t _ {m}\right)$$ $$V t _ {m} - \frac {g t _ {m} ^ {2}}{2} = V (t _ {m} - \Delta t) - \frac {g (t _ {m} - \Delta t) ^ {2}}{2}$$ $$V t _ {m} - \frac {g t _ {m} ^ {2}}{2} = V t _ {m} - V \Delta t - \frac {g t _ {m} ^ {2}}{2} + g t _ {m} \Delta t - \frac {g \Delta t ^ {2}}{2}$$ $$g t _ {m} \Delta t = V \Delta t + \frac {g \Delta t ^ {2}}{2}$$ $$2 g t _ {m} \Delta t = 2 V \Delta t + g \Delta t ^ {2}$$ $$t _ {m} = \frac {\Delta t (2 V + g)}{2 g \Delta t} = \frac {4 s \cdot \left(2 \cdot 9 8 \frac {m}{s} + 9 . 8 \frac {m}{s ^ {2}} \cdot 4 s\right)}{2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 4 s} = 1 2 s$$

Answer: bodies will meet after 12s.