Can a Cannonball's Height Exceed Its Range?

[poseidon007]

Homework Statement

A cannonball is launched with initial velocity of magnitude v₀ over a horizontal surface. At what minimum angle θ_min 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) θ_min = 76◦
(B) θ_min = 72◦
(C) θ_min = 60◦
(D) θ_min = 45◦
(E) There is no such angle, as R > H for all range problems.

Homework Equations

d = (v_i+v_f)/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.

 

[TSny]

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.]

 

[TSny]

d = (v_i+v_f)/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?

 

[poseidon007]

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!

 

[TSny]

Good job.