A ball is dropped from the top of a tower. After $2 s$ another ball thrown vertically downwards with a speed of $40 m s^{- 1}$ , After how much time and at what distance below the the top of tower the balls meet?

Question

Verified by Toppr

Answer 1

If $2$ nd ball takes $t$ time before meeting .
ball $1$ has traveled for $(t + 2) s$
Both balls traveled same dist.
$∴ ∣ d f r a c 12 (t + 2)^{2} = u t + \frac{1}{2} g^{(t)^{2}}$
$\frac{1}{2} \times 10 (t + 2)^{2} = 40 t + \frac{t}{2} \times 10 (t)^{2}$
$5 (t + 2)^{2} = 40 t + 5 6^{2}$
$5 (t^{2} + 4 + 4 t) = 40 t + 5 t^{2}$
$h = \frac{1}{2} g t^{2}$ for $1$ st ball
$h = 40 (t - 2) + \frac{1}{2} g (t - 2)^{2}$ $2$ nd ball
$\frac{1}{2} g t^{2} = 40 (t - 2) + \frac{1}{2} (t - 2)^{2}$
$40 (t - 2) = \frac{1}{2} g (t^{2} - (t - 2))^{2}$
$4 t - 8 = 2 t - 2 \Rightarrow t = 3$
$h = \frac{1}{2} g t^{2}$
$= \frac{1}{2} \times 10 \times (3)^{2} = 45 m$

Answer 2

If

2

nd ball takes

t

time before meeting .
ball

1

has traveled for

(t + 2) s

Both balls traveled same dist.

∴ ∣ d f r a c 12 (t + 2)^{2} = u t + \frac{1}{2} g^{(t)^{2}}

\frac{1}{2} \times 10 (t + 2)^{2} = 40 t + \frac{t}{2} \times 10 (t)^{2}

5 (t + 2)^{2} = 40 t + 5 6^{2}

5 (t^{2} + 4 + 4 t) = 40 t + 5 t^{2}

h = \frac{1}{2} g t^{2}

for

1

st ball

h = 40 (t - 2) + \frac{1}{2} g (t - 2)^{2}

2

nd ball

\frac{1}{2} g t^{2} = 40 (t - 2) + \frac{1}{2} (t - 2)^{2}

40 (t - 2) = \frac{1}{2} g (t^{2} - (t - 2))^{2}

4 t - 8 = 2 t - 2 \Rightarrow t = 3

h = \frac{1}{2} g t^{2}

= \frac{1}{2} \times 10 \times (3)^{2} = 45 m