Question

The horizontal range of a bullet fired with angle of projection 45° to the horizontal is 360 m . If it is fired from lorry moving in the direction of bullet with the uniform velocity 18 km/h and with same elevation , what is the new range horizontal distance traveled by the bullet ? ( Take g = 10 m/s² )

Ans -: 402.42 m

Answer 1

Given that when θ = 45°, horizontal range, R=360 m.

Let u be the velocity with which the bullet is fired and v be the velocity of the lorry.

refer attached 1st pic...

we get,

u = 59.4 m/s

Now,

Given that the lorry is moving with a uniform velocity,

v= 18 km/ hr =(18×5/18)m/s = 5 m/s

In horizontal direction, as the motion is uniform the acceleration is zero.

So we can write the displacement in the horizontal direction as
R = u(x) × T

Here u(x) = u cos θ+ v

        u(y) = u sin θ

Hence Horizontal range

R = (u cosθ+v)× T -------(1)

for solving equation refer attached 2nd pic...

The horizontal distance travelled by the bullet = 402.9 m

HOPE THIS ANSWER WILL HELP U...

Answer 2

Answer is in attachment

thank you