Question

A cannon fires a cannonball with an initial velocity of 300 m/s at 58.0° above the horizontal. The cannonball impacts a raised plateau 37.0 s after firing. Let the +x-direction be directly ahead of the cannon and the +y-direction be upward. Find the x- and y-coordinates of the cannonball's impact point, relative to its firing point (in m).

x = _________m

y = _________m

Answer 1

Answer 2

We can use the equations of motion to find the x- and y-coordinates of the cannonball's impact point.

In the x-direction: The cannonball has a constant horizontal velocity of 300 cos(58°) m/s. Therefore, the x-distance traveled by the cannonball is:

x = (300 cos(58°))(37.0 s) = 8.55 × 10^4 m

In the y-direction: The initial vertical velocity of the cannonball is 300 sin(58°) m/s, and the acceleration due to gravity is -9.81 m/s^2. Therefore, the vertical distance traveled by the cannonball can be found using the following kinematic equation:

y = y0 + v0y t + (1/2)at^2

where y0 = 0 (since the cannonball is fired from ground level), v0y = 300 sin(58°) m/s, and t = 37.0 s. Plugging in these values gives:

y = (300 sin(58°))(37.0 s) + (1/2)(-9.81 m/s^2)(37.0 s)^2 = 5.09 × 10^4 m

Therefore, the x-coordinate of the impact point is 8.55 × 10^4 m and the y-coordinate is 5.09 × 10^4 m, relative to the firing point.