Question

A cannonball is fired in the air at an angle of 45°. How far does it travel before it is 600 feet above ground? (Assume the cannonball travels in a straight line. Ignore the force of gravity and wind resistance. Round your answer to the nearest foot.)

Answer 1

Explanation:

Since the cannonball is fired at an angle of 45°, it will have the same horizontal and vertical velocities.

Let x be the distance it travels before it is 600 feet above ground. At this point, its vertical displacement will be 600 feet, and we can use the formula for vertical displacement:

y = v0*t + (1/2)at^2

Since there is no initial vertical velocity and we are ignoring gravity, a = 0 and the equation simplifies to:

y = 0*t + 0 = 600

Solving for t, we get:

t = sqrt(2y/a) = sqrt(2*600/0) = undefined

This means that the cannonball will never be 600 feet above ground if we ignore gravity. Since this is not realistic, we cannot answer this question without additional information or by assuming that gravity is acting on the cannonball.