
A bullet shot from a rifle at $25m$ range strikes the target at a point $4.9cm$ below the point at which the barrel is pointing horizontally. The muzzle velocity of the bullet is
A. ${\text{50m}}{{\text{s}}^{ - 1}}$
B. ${\text{100m}}{{\text{s}}^{ - 1}}$
C. ${\text{125m}}{{\text{s}}^{ - 1}}$
D. ${\text{250m}}{{\text{s}}^{ - 1}}$
Answer
568.5k+ views
Hint: Before proceeding with the solution, first convert all the given terms in the same system of units. Then use the second equation of motion in the vertical direction - $S = ut + \dfrac{1}{2}g{t^2}$ and the velocity that is distance travelled per time taken. Substitute known values and find the unknown terms.
Complete step by step answer:
Distance travelled by the bullet is \[d = 25m\]
Bullet strikes the target at a point, $s = 4.9cm = 0.049m$
Gravitational acceleration, $g = 9.8m/{s^2}$
Initial velocity, $u = 0m/s$
Now, according to the second equation of the motion in the vertical direction –
$S = ut + \dfrac{1}{2}g{t^2}$
Place the known values in the above equation and make unknown “t” the subject –
$0.049 = 0 + \dfrac{1}{2} \times 9.8 \times {t^2}$
Simplify the above equation –
\[
0.049 = 4.9{t^2} \\
{t^2} = \dfrac{{0.049}}{{4.9}} \\
\implies {t^2} = \dfrac{{0049}}{{1000}} \times \dfrac{{10}}{{49}} \\
\implies {t^2} = \dfrac{1}{{100}} \\
\implies t = \dfrac{1}{{10}} \\
\therefore t = 0.1s \\
\]
Now, velocity is distance per time taken.
Therefore, the muzzle velocity of the bullet is
$v = \dfrac{d}{t}$
Place the known values in the above equation –
$\implies v = \dfrac{{25}}{{0.1}} \\$
$\therefore v = 250m/s \\ $
Therefore, the required answer - The muzzle velocity of the bullet is $250m/s$
So, the correct answer is “Option D”.
Note:
Remember basic formulas for velocity and its laws of equations to solve these types of sums. Also, always double check the given units in all the terms and the units of the answer required. All the terms should have the same format of units. There are three systems of units.
- MKS unit (Meter Kilogram Second)
- CGS unit (Centimetre gram Second)
- SI unit (System International)
Complete step by step answer:
Distance travelled by the bullet is \[d = 25m\]
Bullet strikes the target at a point, $s = 4.9cm = 0.049m$
Gravitational acceleration, $g = 9.8m/{s^2}$
Initial velocity, $u = 0m/s$
Now, according to the second equation of the motion in the vertical direction –
$S = ut + \dfrac{1}{2}g{t^2}$
Place the known values in the above equation and make unknown “t” the subject –
$0.049 = 0 + \dfrac{1}{2} \times 9.8 \times {t^2}$
Simplify the above equation –
\[
0.049 = 4.9{t^2} \\
{t^2} = \dfrac{{0.049}}{{4.9}} \\
\implies {t^2} = \dfrac{{0049}}{{1000}} \times \dfrac{{10}}{{49}} \\
\implies {t^2} = \dfrac{1}{{100}} \\
\implies t = \dfrac{1}{{10}} \\
\therefore t = 0.1s \\
\]
Now, velocity is distance per time taken.
Therefore, the muzzle velocity of the bullet is
$v = \dfrac{d}{t}$
Place the known values in the above equation –
$\implies v = \dfrac{{25}}{{0.1}} \\$
$\therefore v = 250m/s \\ $
Therefore, the required answer - The muzzle velocity of the bullet is $250m/s$
So, the correct answer is “Option D”.
Note:
Remember basic formulas for velocity and its laws of equations to solve these types of sums. Also, always double check the given units in all the terms and the units of the answer required. All the terms should have the same format of units. There are three systems of units.
- MKS unit (Meter Kilogram Second)
- CGS unit (Centimetre gram Second)
- SI unit (System International)
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