A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground? Show all work.
find the time of flight (t) ... 1.5 = 1/2 g t^2
(travel distance) = (time of flight) * (forward velocity)
To solve this problem, we need to analyze the motion of the cannonball in two dimensions: horizontally and vertically.
Let's start by considering the vertically downward motion of the cannonball. We can use the equation of motion:
h = ut + (1/2)gt^2
where h is the height, u is the initial vertical velocity, g is the acceleration due to gravity, and t is the time.
Since the cannonball is fired horizontally, its initial vertical velocity is zero (u = 0). The height of the cannonball is given as 1.5 m (h = 1.5 m). The acceleration due to gravity is approximately 9.8 m/s^2 (g = 9.8 m/s^2). We need to find the time it takes for the cannonball to reach the ground (t).
Using the equation, we can rearrange it to solve for t:
t = √(2h/g)
Plugging in the given values:
t = √(2 * 1.5 / 9.8)
t ≈ √(0.30612245)
t ≈ 0.553 s (rounded to three decimal places)
Now that we have the time it takes for the cannonball to reach the ground, we can calculate the horizontal distance it travels.
The horizontal distance is given by the equation:
d = v * t
where d is the distance, v is the initial horizontal velocity, and t is the time.
The initial horizontal velocity is given as 48.1 m/s (v = 48.1 m/s) and the time is 0.553 s (t ≈ 0.553 s). Substituting these values into the equation:
d = 48.1 * 0.553
d ≈ 26.6263 m (rounded to four decimal places)
Therefore, the cannonball would travel approximately 26.6263 meters horizontally before landing on the ground.