Question : A gunner decided to hit a suspected car running on a road. The gun kept on a straight horizontal road is used to hit a car travelling along the same road away from the gun with uniform velocity of 72 km/h. The car is at a distance of 500 m from the gun when the gun is fired at an angle of 45° with the horizontal. [Case study bases question]
i) The distance of the car from the gun when the bullet hit the car is
a) 600 m
b) 650 m
c) 644 m
d) 744 m ⭐
ii) The maximum height attained by the bullet during its flight is:
a) 180 m
b) 183 m
c) 186 m ⭐
d) 193.8 m
iii) Time taken by the bullet to hit the car is:
a) 12.98 s
b) 18.014 s
c) 18.140 s
d) 12 s ⭐
iv) The distance travelled by the car before it was struck by the bullet
a) 160 m
b) 170 m
c) 244 m ⭐
d) 200 m
v) The speed of projection of the bullet from the gun is :
a) 86.00 m/s
b) 86.56 m/s
c) 86.25 m/s ⭐
d) 86.05 m/s
Doubt by Muskan
Solution :
Uniform velocity of the car, v = 72 km/h
=72×(5/18) = 20 m/s
=72×(5/18) = 20 m/s
Let the initial velocity with which the bullet is fired be = u
Angle of projection (θ) = 45°
so
ux=ucosθ
uy=usinθ
OA'=500 m
OA = ?
OA = ?
Let T = Time of flight
and R = Range of the Projectile
We know,
R =u2sin2θ/g
R =u2sin2θ/g
OA'+A'A = u2sin(2×45°)/g
500 +20(T) = u2/g [∵Distance = Speed × Time]
500 +20(T) = u2/9.8 — (1)
500 +20(T) = u2/g [∵Distance = Speed × Time]
500 +20(T) = u2/9.8 — (1)
Now, Time of flight
T = 2usinθ/g
T=2usin45°/9.8
T = 2usinθ/g
T=2usin45°/9.8
T = 2u/9.8√2
T=√2u/9.8 — (2)
T=√2u/9.8 — (2)
Putting in eq (1)
500+20(√2u/9.8) = u2/9.8
u2/9.8-20√2u/9.8-500=0
u2-20√2u-500×9.8 = 0
u2/9.8-20√2u/9.8-500=0
u2-20√2u-500×9.8 = 0
u2-20√2u-4900=0
Solving this quadratic equation
a=1
b=-20√2
c=-4900
D=b2-4ac
D=(-20√2)2-4(1)(-4900)
D=800+19600
D=20400
D=√(2×2×10×10×51)
D=20√51
D=20400
D=√(2×2×10×10×51)
D=20√51
u = [-b±√D]/2a
u = [-(-20√2)±20√51]/(2×1)
u = [20√2±20√51]/2
u = 20[√2±√51]/2
u = 10[√2±√51]
u = 10[1.414±7.141]
Neglecting the negative value of u.
u = [-(-20√2)±20√51]/(2×1)
u = [20√2±20√51]/2
u = 20[√2±√51]/2
u = 10[√2±√51]
u = 10[1.414±7.141]
Neglecting the negative value of u.
u = 10[1.414+7.141]
u = 10[8.555]
u = 85.56 m/s
u = 10[8.555]
u = 85.56 m/s
R=u2/9.8
R =(85.56)2/9.8
R = 746.99 m
R = 746.99 m
i) The distance of the car from the gun when the bullet hit the car will be equal to the to the range of the projectile which is equal to 746.99 m
ii) Maximum Height attained by the bullet
H=u2sin2θ/2g
H=(85.56)2sin2(45°)/(2×9.8)
H=7320.51×(1/2)/19.6
H=3660.255/19.6
H=186.75
H=u2sin2θ/2g
H=(85.56)2sin2(45°)/(2×9.8)
H=7320.51×(1/2)/19.6
H=3660.255/19.6
H=186.75
So, the nearest answer would be c) 186 m.
iii) Time taken by the bullet to hit the car = Time of flight
Using Eq (2)
T=√2u/9.8
T=√2×85.56/9.8
T=1.414×85.56/9.8
T=120.98184/9.8
T=12.34508
T≈12.34 sec
Using Eq (2)
T=√2u/9.8
T=√2×85.56/9.8
T=1.414×85.56/9.8
T=120.98184/9.8
T=12.34508
T≈12.34 sec
So, the nearest answer is a) 12.98 s.
iv) Distance travelled by the car before it was struck by the bullet = A'A = OA-OA'
= R-OA'
= R-OA'
= 746.99 - 500
= 246.99 m
v) speed of projection of the bullet from the gun = u
u = 85.56 m/s