Question

Two paper screen A and B are separated by a distance of 100m. A bullet A and then B. The hole in B is 10cm below the hole in A.If the bullet is travelling horizontally at the time of hitting the screen A, calculate the velocity of the bullet when it hits the screen

Answer 1

Answer:

700 m/s

Explanation:

At screen A the bullet has no vertical velocity. Suppose it takes time “t” to reach the screen B. It falls by a height of 10cm = 0.1 m by the time it reaches the screen B.

Therefore,

0.5 gt^2 =0.1

( g= 9.8m/s^2)

t^2=1/49

t=1/7 sec

Suppose the horizontal velocity of the bullet at screen A is V. The bullet covers the horizontal distance of 100m between screen A and B in “t” sec.

Therefore,

Vt=100

As calculated above, t=1/7 sec

Therefore,

Vx 1/7=100

V= 7x 100

V=700 m/s

Answer 2

Answer:

Explanation:

Let x = 100 m and h = 10 cm

Using equation of motion we can then calculate the time of the motion as, 

h = ut + 1/2 gt²

So we will then assume initial velocity to be zero and g = 10 m/s² 

We will get 

t = √2h/g

t = √2 x 10 cm/ 10 m/s²

 t = √2 x 0.1 m / 10m/s²

t = 0.14 s 

Now again we will use equation of motion as, 

x = vt or v = x/t

we get, 

v = 100 m/0.14 s

v = 707.11m/s