A cannonball is launched with initial velocity of magnitude v0 over a horizontal surface. At what minimum angle
θmin above the horizontal should the cannonball be launched so that it rises to a height H which is larger than the
horizontal distance R that it will travel when it returns to the ground?
(A) θmin = 76◦
(B) θmin = 72◦
(C) θmin = 60◦
(D) θmin = 45◦
(E) There is no such angle, as R > H for all range problems.

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Netta00 Netta00 · Answer 1 · 2019-10-02T14:07:53+02:00

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

[tex]R=\dfrac{V_o^2sin2\theta }{g}[/tex]

Height h

[tex]h=\dfrac{V_o^2sin^2\theta }{2g}[/tex]

Given that h should be larger than R

h > R

[tex]\dfrac{V_o^2sin^2\theta }{2g}>\dfrac{V_o^2sin2\theta }{g}[/tex]

[tex]{sin^2\theta }>2{sin2\theta }[/tex]

[tex]{sin^2\theta }>4{sin\theta\ cos\theta }[/tex]

[tex]{tan\theta }>4[/tex]

θ >75.96°

So minimum angle should be 75.96° or 76°.