Question

A gun kept on a straight horizontal road is used to hit a car traveling along the same road away fron the gun with a uniform speed of 72*21/2 km/h. The car is at a distance of 50m from the gun, when the gun is fired at an angle of 45o with the horizontal. Find
a) the distance from the gun from the car when the shell hits it.
b)the speed of projection of shell from the gun .[g=10m/s2]
Please explain each and every step.
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Answer 1

Correction(By Looking at other sources) : Speed of gun : $72 \sqrt{2} /: km/h$


Final Answer :
(a) 250m
(b) 50 m/s

Steps:
1) Let 't' be the time when gun hits the car after moving from 50 m distance from gun.
We observe that, if gun hits the car then
t = Time period of projectile of bullet from gun.
So, Initial speed of bullet be 'v'.
$t = \frac{2v \sin( \theta) }{g}$

2) We also observe that,
Speed of car, u = 72√2 km/h = 20√2 m/s

Distance covered by car
Range of Projectile = 50m + 20√2 *t
$= > \frac{ {v}^{2} \sin(2 \times 45 \degree) }{g} = 50 + 20 \sqrt{2} (\frac{2v \: \sin(45 \degree) }{g} ) \\ = > {v}^{2} - 40v - 500 = 0 \\ = > (v - 50)(v + 10) = 0 \\ = > v = 50 \: \: \: (v \neq - 10)$


(b) Hence, Speed of Projection of shell
=50m/s

(a) Distance from the gun :
Range of Projectile :
$= \frac{ {v}^{2} \sin(2 \times 45 \degree) }{g} = \frac{ {50}^{2} \times 1}{10} = 250 \: m$

Hence, Distance from gun = 250 m

Answer 2

Done!!

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