Question

The maximam range of a bullet fired from a toy pistol mounted on a car at rest is R=40m. What will be the acute angle of inclination of the pistol for maximum range when car is moving in the direction of firing with uniform velocity v=20m/s, on the horizontal surface?​

Answer 1

Maximum range of projectile is given by formula

$r \: = \frac{ {v}^{2} }{g}$

now as per the formula

$40 \: = \: \frac{ {v}^{2} }{10} \\ v \: = \: 20 \: m {s}^{ - 1}$

with respect to ground the velocity must make an angle of 45 degree

so its velocity components along X and Y direction must be

$vx \: = 10 \sqrt{2 } \\ vy = 10 \sqrt{2}$

velocity with respect to car is given as

$vx' = 10 \sqrt{2} - 20 = - 5.86ms^{ - 1}$

now in order to find the inclination

$\alpha = \tan^{ - 1} \frac{vy}{vx } \\ \: \: \: \: = \frac{10 \sqrt{2} }{5.86} \\ \: \: \: \: = 67.5 \: degrees \:$

so angle of inclination must be 67.5°.

Hope it helps you

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Answer 2

Given

taken g = 10m/s^2

Rmax = 40m

u = 20m/s

R = u^2 sin2@/g

40= 400sin2@/10

sin2@ = 1

2@ = sin^-1sinπ/2

@ = π/4

@ =45°

since any body When throw at an inclination of 45° the range followed by body is always maximum

from above it is too proved .

Concept

the pistol is attached to car at rest so the velocity of car is equal to gun pistol velocity .

there for

u (initial velocity ) = 2m/s

prove of formula

we assume that a body follow a projectile motion having initial velocity u ,angle be @ ,time is t ,and x be the horizontal direction motion .

Now

we know that

T = 2usin@)/g

x =( ucos@ )t

so

R = u(2usin@cos@)/g

R = 2u^2 sin@cos@/g

R = u^2 sin2@/g