Can a Cannonball's Height Exceed Its Range?

• poseidon007
In summary, the question asks for the minimum launch angle for a cannonball to reach a height larger than the horizontal distance it travels when returning to the ground. Using the equations for distance and time, the minimum angle is found to be 75.9°.
poseidon007

Homework Statement

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.

d = (vi+vf)/2)*t

The Attempt at a Solution

H = (1/2)(v0sinθ)(t) and R = (v0cosθ)(t)

Thus, if H = R, then (1/2)(v0sinθ)(t) = (v0cosθ)(t)
=>tanθ = 2, so θ = 63.4°. I'm probably making a really obvious mistake here, but I'm not seeing it. Any help would be appreciated.

Do you need to include the effect of gravity somewhere? [EDIT: Nevermind, you are using the equation d = (vi + vf)*t/2 which doesn't require knowing the acceleration.]

Last edited:
poseidon007 said:
d = (vi+vf)/2)*t

H = (1/2)(v0sinθ)(t) and R = (v0cosθ)(t)

Thus, if H = R, then (1/2)(v0sinθ)(t) = (v0cosθ)(t)
=>tanθ = 2, so θ = 63.4°.
I think I now see what you're doing. Does the t in the H equation represent the same time as the t in the R equation?

ohhhh i see now. the t in the H equation is the time to get to maximum height, which is half of the t in the R equation. so it would really be (1/2)(v0sinθ)(t/2) = (v0cosθ)(t) which gives tanθ = 4 so θ = 75.9° (A). Thanks so much!

Good job.

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