Question

A rifle is aimed horizontally at the center of a large target 50.4 m away. If the distance from the center of the target to the point where the bullet strikes the target is 1.51 cm, calculate the initial speed of the bullet in m/s?

Answer 1

Distance of the target, S = 50.4 m

Also, it is given that bullet hits the point which is 1.51 cm below the target. So we would denote this distance as y.

i.e y = 1.51 cm = 0.0151 m

(Because 1 cm = 0.01 m; so 1.51 cm = 0.01*1.51 = 0.0151 m )

Now, let us first calculate the time taken by the bullet to cover this vertical distance y.

i.e y = ut+ gt² / 2 ....eqn 1

where g is the acceleration due to gravity and has value = 9.8 m/s^{​​​​​​2}

Since, u= 0

Hence, we can re-write the eqn 1 as

y = 0*t + gt^{​​​​​2}/2 = gt²/2

Putting value of y and g in this eqn, we get

0.0151= 9.8*t^{​​​​​​2}​​​​​/2

0.0151*2/9.8 = t^{​​​​​​2}​​​​​

0.0030 = t^{​​​​2}

or t = (0.0030)^1/2 = 0.055 s

i.e t = 0.055 seconds which is the time taken by the bullet to cover vertical distance of y = 0.0151 m below the target.

We would now write the formula for horizontal distance S traveled by the bullet in this time interval i.e

S = vt

Where v is the velocity of the bullet when it left the rifle.

Hence v = S/t

We know, that S = 50.4 m and t = 0.055 seconds

Hence, v = 50.4/0.055 = 916.36 m/s

i.e velocity of bullet when it left the rifle is v = 916.36 m/s.