Question : Assume that a bullet P is fired from a gun when the angle of elevation of the gun is 30°. Another bullet Q is fired from the gun when the angle of elevation is 60°. If the vertical height attained in the 2nd case is x times the vertical height attained in 1st case, then find the value of x.
Doubt by Jaskirat
Solution :
H2=xH1 (Given)
Maximum Height of Angular Projectile is given by
H=u2sin2θ/2g
Initial velocity u of the bullet remain same in both the cases because it is being from the same gun.
Case 1 :
θ1 = 30°
H1=u2sin2θ1/2g
H1=u2sin230°/2g
H1=u2(1/2)2/2g
H1=u2/8g — (2)
H1=u2sin230°/2g
H1=u2(1/2)2/2g
H1=u2/8g — (2)
Case 2 :
θ2 = 60°
H2=u2sin2θ2/2g
H2=u2sin260°/2g
H2=u2(√3/2)2/2g
H2=3u2/8g — (2)
H2=u2sin260°/2g
H2=u2(√3/2)2/2g
H2=3u2/8g — (2)
Dividing equation (2) by equation (1)
H2/H1 = [3u2/8g]/[u2/8g]
H2/H1 = 3
x = 3
x = 3
Hence, the required value of x is 3.