Question

A cannonball is launched with initial velocity of magnitude v0 over a horizontal surface. At what minimum angle

Answer 1

Answer:

A)   θmin= 76°.

Explanation:

Given that

Speed of cannonball = Vo

Lets take θ is the angle measure from horizontal

We know that

Range R

$R=\dfrac{V_o^2sin2\theta }{g}$

Height h

$h=\dfrac{V_o^2sin^2\theta }{2g}$

Given that h should be larger than R

         h > R

$\dfrac{V_o^2sin^2\theta }{2g}>\dfrac{V_o^2sin2\theta }{g}$

${sin^2\theta }>2{sin2\theta }$

${sin^2\theta }>4{sin\theta\ cos\theta }$

${tan\theta }>4$

        θ  >75.96°

So minimum angle should be 75.96° or 76°.