Question

A wooden block of mass 10 g is dropped from the top of a cliff 100 m high. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff upward with a velocity of 100 m/s. At what time will the bullet and the block meet?

• A. 1s
• B. 7s
• C. 0.5s
• D. 10s

Answer 1

The correct option is A

1s

Step 1: Given data:

1. A wooden block of mass 10 g is dropped from the top of a cliff 100 m high.
2. Simultaneously, a bullet of mass 10 g is fired from the foot of the cliff upward with a velocity of 100 m/s.

From the equations of motion, we have

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

Here,

s is the distance

u is the initial velocity

a is the acceleration

t is the time

Let x and x' be the distance covered by the block and bullet respectively before the collision.

Then, $x+x\text{'}=100m$

Step 2: Calculating distance for the block:

$$\begin{gathered} u=0 \\ a=g \\ x=0+\frac{1}{2}(9.8)t^2=4.9t^2 \end{gathered}$$

Step 3: Calculating distance for the bullet:

$$\begin{gathered} u=100m/s \\ a=g=-9.8m{s}^{-2}(astheaccelerationisinoppositedirection) \\ x\text{'}=ut-\frac{1}{2}g{t}^2 \\ x\text{'}=100t-\frac{1}{2}(9.8)t^2=100t-4.9t^2 \end{gathered}$$

Step 4: Finding the time:

$$\begin{gathered} Now,x+x\text{'}=100 \\ 100t-4.9t^2+4.9t^2=100 \\ 100t=100 \\ t=1sec \end{gathered}$$

Therefore, the time will be 1 sec.

Hence, the correct option is (A).