Question

Q1. A bell is thrown upward with a velocity of 29.4m/s. After 3 seconds another ball is thrown upwards with a velocity of 19.6m/s . at what height and at what time will the two balls collide?
Q2. A stone after falling freely under gravity for 1 second strikes a pane of glass held horizontally. In breaking through the pane it loses half of its velocity. How far will it travel or fall in the next 2 seconds?

Ans1.. 2.25 seconds after throwing the second stone and at 19.6 height
Ans2.. It will travel or fall till 29.4 meters in the next 2 seconds.

<<<<<<Please Help>>>>>>

Answer 1

Q1. A bell is thrown upward with a velocity of $29.4\mathrm{m/s}$. After 3 seconds another ball is thrown upwards with a velocity of $19.6\mathrm{m/s}$. At what height and at what time will the two balls collide?

Q2. A stone after falling freely under gravity for 1 second strikes a pane of glass held horizontally. In breaking through the pane it loses half of its velocity. How far will it travel or fall in the next 2 seconds?

Solution

Q1

$$h_1 = v_1 t - \frac{g t^2}{2}, \quad h_2 = v_2 (t - 3) - \frac{g (t - 3)^2}{2}$$ $$h_1 = h_2 \gg v_1 t - \frac{g t^2}{2} = v_2 (t - 3) - \frac{g (t - 3)^2}{2} \gg$$ $$t = \frac{3 v_2 + 4.5 g}{v_2 - v_1 + 3 g} = 5.25 \, \text{s}$$

or $(t - 3) = 5.25 - 3 = 2.25 \, \text{s}$ after throwing the second stone

$$h(2.25) = 29.4 \times 5.25 - \frac{9.8 \times 5.25^2}{2} = 19.3 \, \text{m}$$

Q2

Velocity of the stone flew one second

$$v = g t = g \cdot 1 = g$$

In breaking through the pane it loses half of its velocity, so

$$v_0 = v / 2 = g / 2$$ $$h = v_0 t + \frac{g t^2}{2} = \frac{g}{2} t + \frac{g t^2}{2} \gg h(2) = \frac{g}{2} t + \frac{g t^2}{2} = 3 g = 29.4 \, \text{m}$$