A cannon is fired horizontally from a top of a 15-m platform. What is the cannon balls horizontal speed when it hits the ground 250 m downrange?
The time (t) that it takes to hit the ground is determined by the height above ground. For horizontal firing,
t = sqrt(2H/g) = 1.749 s
The horizontal speed must be (at all times)
250/1.749 = 143 m/s
To find the horizontal speed of the cannonball when it hits the ground, we can use the formula for horizontal distance traveled:
d = v * t
Where:
d = horizontal distance traveled (250 m)
v = horizontal speed of the cannonball
t = time of flight
We have the distance and we can calculate the time of flight using the formula for vertical motion:
h = (1/2) * g * t^2
Where:
h = height of the platform (15 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight
First, let's calculate the time of flight:
15 = (1/2) * 9.8 * t^2
Rearranging the equation:
t^2 = (15 * 2) / 9.8
t^2 = 30 / 9.8
t^2 = 3.06
Taking the square root of both sides:
t = √3.06
t ≈ 1.75 seconds
Now we can substitute the time of flight back into the equation for horizontal distance traveled to find the horizontal speed:
250 = v * 1.75
Dividing both sides by 1.75:
v ≈ 142.86 m/s
Therefore, the cannonball's horizontal speed when it hits the ground 250 m downrange is approximately 142.86 m/s.
To determine the cannonball's horizontal speed when it hits the ground, we can use the principle of projectile motion. When a cannonball is fired horizontally, its vertical motion is influenced by gravity, while its horizontal motion remains constant.
First, let's analyze the vertical motion of the cannonball. We know that the cannonball's initial vertical position is 15 m above the ground, and it falls downward due to gravity. The vertical displacement can be determined using the equation:
Δy = v₀y * t + (1/2) * g * t²
Since the initial vertical velocity (v₀y) is 0 (as the cannonball is fired horizontally), we can simplify the equation:
Δy = (1/2) * g * t²
where Δy is the vertical displacement, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time of flight.
Since the cannonball hits the ground 250 m downrange, the horizontal distance traveled (range) is equal to the initial horizontal velocity (v₀x) multiplied by the time of flight (t):
Range = v₀x * t
We want to find the horizontal speed, which is the magnitude of the horizontal velocity (v₀x). To do that, we need to find the value of t, the time of flight.
Using the formula for time of flight of a projectile:
t = sqrt(2 * Δy / g)
we can substitute the value of Δy (which is the height of the platform, 15 m) and g (acceleration due to gravity, 9.8 m/s²) into the equation:
t = sqrt(2 * 15 / 9.8)
Simplifying, we find:
t ≈ 1.32 seconds
Now that we have the time of flight, we can find the horizontal speed (v₀x) using the range formula:
Range = v₀x * t
250 m = v₀x * 1.32 seconds
Simplifying, we can solve for v₀x:
v₀x ≈ 189.4 m/s
Therefore, the cannonball's horizontal speed when it hits the ground 250 m downrange is approximately 189.4 m/s.