Question

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.

Answer 1

A) θmin= 76°.

Step-by-step explanation:

Given that

Speed of cannonball = Vo

Lets take θ is the angle measure from horizontal

We know that

Range R

`R=(V_o^2sin2\theta )/(g)`

Height h

`h=(V_o^2sin^2\theta )/(2g)`

Given that h should be larger than R

h > R

`(V_o^2sin^2\theta )/(2g)>(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°.