Question

A wooden block of mass $10g$ is dropped from the top of a cliff $100m$ high. Simultaneously, a bullet of mass $10g$ is fired from the foot of the cliff upwards with a velocity $100m{s}^{-1}$ . After what time, does the bullet and the block meet?

Answer 1

Step 1: Given and assumption,

Let, the bullet and block will meet at the distance $x$ from the ground, and the acceleration is the acceleration due to gravity.

Downward direction is taken as positive.

The initial velocity of the block, $u=0m{s}^{-1}$ (Dropped)

Distance traveled by the block, $s=\left(100-x\right)m$

The initial velocity of the bullet, $u=-100m{s}^{-1}$ (Thrown upwards)

Distance traveled by the bullet, $s=\left(-x\right)m$

Step 2: Formula

According to the second equation of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, $s$ is distance, $u$ is initial velocity, $t$ is time and $a$ is acceleration.

Step 3: Derive the equation for block and bullet,

Substituting and making equations for bullet and block,

The second equation of motion for block,

Substituting $s=\left(100-x\right)m$and $u=0m{s}^{-1}$ in $s=ut+\frac{1}{2}a{t}^2$,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow 100-x=\left(0\right)t+\frac{1}{2}g{t}^2$

$\Rightarrow 100-x=\frac{1}{2}g{t}^2$

The second equation of motion for the bullet,

Substituting $u=-100m{s}^{-1}$ and $s=\left(-x\right)m$ in $s=ut+\frac{1}{2}a{t}^2$,

$s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow -x=(-100)t+\frac{1}{2}g{t}^2$

$\Rightarrow -x+100t=\frac{1}{2}g{t}^2$

Step 4: Calculating the time,

From the equations of bullet and block, equating the values of $\frac{1}{2}g{t}^2$ ,

$\Rightarrow -x+100t=100-x$

Solving to calculate the time,

$\Rightarrow 100t=100$

$\Rightarrow t=\frac{100}{100}=1sec$

Hence, at the time, $t=1sec$ bullet and block will meet.