Question

Mary wants to throw a can straight up in air and then hits it with second can.she the collision to occur at height h=10m above the throw point. In addition,she knows that she needs t1=4.0s between successive throws. Assume that she throws both cans with same speed.Take g=9.81m/^2. How long it takes after the first can has been thrown into air for the two cans to collide?

Answer 1

Question #79756, Physics / Classical Mechanics

Mary wants to throw a can straight up in air and then hits it with second can. She the collision to occur at height $h = 10m$ above the throw point. In addition, she knows that she needs $t_1 = 4.0s$ between successive throws. Assume that she throws both cans with same speed. Take $g = 9.81m/s^2$. How long it takes after the first can has been thrown into air for the two cans to collide?

Solution

Let the height of the first can be $x$, that of the second can be $y$; and both cans be thrown at speed $v$.

$$x(t) = v t - g \frac{t^2}{2}$$ $$y(t) = v (t - t_1) - g \frac{(t - t_1)^2}{2}$$ $$x(t) = y(t) = h$$

From the first height equation:

$$v = \frac{h}{t} + \frac{g t}{2}$$

Substituted into the second equation:

$$h = \left(\frac{h}{t} + \frac{g t}{2}\right) (t - t_1) - \frac{g (t - t_1)^2}{2}$$ $$10 = \left(\frac{10}{t} + 9.81 \frac{t}{2}\right) (t - 4) - \frac{9.81 (t - 4)^2}{2}$$ $$t = 4.46 \, \text{s}.$$

Answer: 4.46 s.

Answer provided by https://www.AssignmentExpert.com