a cannon tilted up at 33 degree angle fires a cannon ball at 88 m/s from atop a 12.0 m high fortress wall. What’s the balls impact speed on the ground?
the height of the ball is
h(t) = 12 + (88 sin33°)t - 4.9t^2
h=0 at t=10
v(t) = 88 sin33° - 9.8t
so plug and chug
alternatively, consider the ball as being thrown down with an initial speed of
-88 sin33° m/s
and find how long it takes to drop 12m.
Plug that into the v(t) equation
To find the impact speed of the cannonball on the ground, we need to consider the projectile motion.
First, let's break down the initial velocity of the cannonball into horizontal and vertical components. The vertical component can be calculated using the initial velocity (88 m/s) and the sine of the launch angle (33 degrees). The horizontal component can be calculated using the initial velocity (88 m/s) and the cosine of the launch angle (33 degrees).
Vertical component of initial velocity (Viy) = Initial velocity (V) * sine (θ)
Viy = 88 m/s * sin(33°)
Viy = 88 m/s * 0.5446
Viy ≈ 47.872 m/s
Horizontal component of initial velocity (Vix) = Initial velocity (V) * cosine (θ)
Vix = 88 m/s * cos(33°)
Vix = 88 m/s * 0.8365
Vix ≈ 73.624 m/s
Now, we can find the time it takes for the cannonball to hit the ground by using the vertical component of the motion. We can use the equation:
Vertical distance (d) = (Viy * t) + (0.5 * g * t^2)
Where d is the vertical distance (12 m), Viy is the vertical component of initial velocity (47.872 m/s), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of flight.
Rearranging the equation, we get:
12 = (47.872 * t) + (0.5 * 9.8 * t^2)
Rearranging further:
4.9t^2 + 47.872t - 12 = 0
Solving this quadratic equation for t, we get two values for time. However, since we're interested in the time it takes for the cannonball to hit the ground, we'll consider the positive value.
Now we can find the time of flight (t) using the quadratic formula:
t = (-b + sqrt(b^2 - 4ac))/(2a)
Using the formula, we find:
t ≈ 1.74 seconds
Finally, we can find the impact speed on the ground using the horizontal component of the motion. We can use the equation:
Horizontal distance (d) = Vix * t
d = 73.624 m/s * 1.74 s
d ≈ 127.756 m
Therefore, the ball's impact speed on the ground is approximately 127.756 m/s.
To find the impact speed of the cannonball on the ground, we need to analyze the projectile motion. We can break the motion into horizontal and vertical components.
Let's start by analyzing the vertical motion:
1. Find the initial vertical velocity (Vy₀): Since the cannonball is fired with an initial velocity of 88 m/s at an angle of 33 degrees above the horizontal, we can find the vertical component of this velocity using the formula Vy₀ = V₀ * sin(θ), where V₀ is the initial velocity and θ is the angle.
Vy₀ = 88 m/s * sin(33°) = 88 m/s * 0.545 = 47.96 m/s
2. Find the time taken to reach the maximum height (t_max): The time taken to reach the peak of the projectile's trajectory can be found using the formula t_max = Vy₀ / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
t_max = 47.96 m/s / 9.8 m/s² = 4.89 s (rounded to two decimal places)
3. Find the maximum height (H): Using the formula H = (Vy₀²) / (2g), we can find the maximum height.
H = (47.96 m/s)² / (2 * 9.8 m/s²) = 117.46 m (rounded to two decimal places)
Next, let's analyze the horizontal motion:
4. Find the time of flight (t_flight): The total time of flight can be calculated using the formula t_flight = 2 * t_max.
t_flight = 2 * 4.89 s = 9.78 s (rounded to two decimal places)
5. Find the horizontal distance traveled (R): Using the formula R = V₀ * cos(θ) * t_flight, where θ is the angle and V₀ is the initial velocity.
R = 88 m/s * cos(33°) * 9.78 s = 689.25 m (rounded to two decimal places)
Finally, let's find the impact speed on the ground:
6. Determine the final vertical velocity (Vy): Since the cannonball hits the ground, the final vertical velocity will be negative and equal to -Vy₀.
Vy = -47.96 m/s
7. Find the impact speed (V_impact): Using the Pythagorean theorem, we can find the impact speed.
V_impact = √(Vx² + Vy²)
V_impact = √((88 m/s * cos(33°))² + (-47.96 m/s)²) = √(374.73² + 2302.87) = √279721.65 = 529.35 m/s (rounded to two decimal places)
Therefore, the ball's impact speed on the ground is approximately 529.35 m/s.