a gun is kept on a straight horizontal road is used to hit a car travelling along the same road away from the gun with the uniform speed of 72*√2 km/h. the car is at a distance of 50m from the gun,when the gun is fired at an angle of 45° with the horizontal. find(1) the distance of the car from the gun when the shell hits it, (2) the speed of projection of the shell from the gun
Answers
Final Answer :
(a) 250m
(b) 50 m/s
2) We also observe that,
Speed of car, u = 72√2 km/h = 20√2 m/s
sin(2×45°)
=50+20
2
(
g
2vsin(45°)
)
=>v
2
−40v−500=0
=>(v−50)(v+10)=0
=>v=50(v≠−10)
(b) Hence, Speed of Projection of shell
=50m/s
(a) Distance from the gun :
Range of Projectile :
sin(2×45°)
=
10
50
2
×1
=250m
Hence, Distance from gun = 250 m
Correction(By Looking at other sources) : Speed of gun :
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'.
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
(b) Hence, Speed of Projection of shell
=50m/s
(a) Distance from the gun :
Range of Projectile :
Hence, Distance from gun = 250 m