Question

2 paper screens A to B are separated by 150 m. A bullet pierces A and then B. The hole in B is 15cm below the hole in A. If the bullet is travelling horizontally at the time of hitting A, then the velocity of the bullet at A is (g = 10ms raised to -2) find velocity of b

Answer 1

Time duration for fall of bullet by 15 cm :
     S = u t + 1/2 at²  =>  0.15 = 0 t + 1/2 * 10 * t²
     So  t = √0.03 sec = 0.1732 sec

Distance traveled horizontally in this time period: 150 meters
Velocity at A is the constant horizontal velocity of the bullet.
         = 150 meters / 0.1732 sec = 866.050 m/sec

Vertical component of velocity of bullet when it reaches paper B
         v = u + at  =>  v = 0 + 10m/sec² * 0.1732 sec = 1.732 m/s

Total Speed of bullet when it pierces paper B :
           V  = √( 866²+1.732² )  =  866.0525 m/sec

Answer 2

Answer:

• 500√3

Explanation:

• Distance between the screens = 150 m

• Height = 15 cm = 0.15 m

`We will take initial velocity, u as 0`

Using, equation of motion :

• s = ut + ½ at²

• { t = √2h/g }

$= > t = \frac{ \sqrt{2 \times 0.15} }{ \sqrt{10}}$

$= >t = \frac{ \sqrt{2 \times 15} }{ \sqrt{1000}}$

$= > t = \frac{ \sqrt{30} }{ \sqrt{1000} }$

$= > t = \frac{ \sqrt{3} }{ \sqrt{100}}$

$= > t = \frac{ \sqrt{3} }{10} \: seconds$

• Distance = velocity × time

• { s = v × t}

• v = s ÷ t

$= >v = \frac{150}{ \sqrt{3} \div 10}$

$= > v = \frac{150 \times 10}{ \sqrt{3} } = \frac{1500}{ \sqrt{3} }$

• By rationalizing :-

$= > \frac{1500 \times \sqrt{3}}{ \sqrt{3}\times\sqrt{3} }$

$= > v \: = \frac{1500 \times \sqrt{3} }{3}$

$= > v = 500 \sqrt{3}$

Hope it helps!

@charlie16