A ship is 690 m away from a port. A shell is fired from the port with initial velocity of vo at an elevation of 45° with the horizontal. The shell hits the ship. Initial velocity of the shell is (Assume ship is stationary and g = 10 m/s²)

Answers 1

Given:

Distance= 690 metres

Angle of elevation= 60°

Gravity (g)= 10 m/s²

To Find:

Initial velocity (V₀)

Solution:

For solving this question, we are going to use the horizontal range formula of projectile motion.

To find the maximum horizontal displacement, we consider the "y" component of the displacement equation to be zero,

⇒y= tanθ. x - [tex]\frac{g}{2u^{2}cos^{2}θ }[/tex]. x²

⇒0= tanθ. x - [tex]\frac{g}{2u^{2}cos^{2}θ }[/tex]. x²

⇒ Range = u²sin2θ/g

[ u= Initial velocity, θ= angle of elevation]

[As all the quantities are in the S.I unit, no conversion is needed.]

Thus, now substituting the given values in the equation, we get,

⇒ 690= [u² × sin (2×45°)] / 10

⇒ 690= [u² × sin 90°] / 10

⇒ 690= u²/10 ∵ [ sin 90°=1]

⇒ 690× 10= u²

⇒ √6900 = u

⇒ u= 83.066 m/sec

Hence, the Initial velocity (V₀) is 83.066 m/sec.