- #1

poseidon007

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## Homework Statement

A cannonball is launched with initial velocity of magnitude v

_{0}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)(v

_{0}sinθ)(t) and R = (v

_{0}cosθ)(t)

Thus, if H = R, then (1/2)(v

_{0}sinθ)(t) = (v

_{0}cosθ)(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.