A cannonball of mass 3.67 kg is shot from a cannon at an angle of 39.51° relative to the horizontal and with an initial speed of 56.41 m/s. As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot?
To find the gain in potential energy at the highest point of the trajectory, we need to calculate the change in height.
The potential energy of an object is given by the equation:
PE = m * g * h
Where:
PE is the potential energy
m is the mass of the object
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height above a reference point
Since the cannonball reaches the highest point, it will have gained potential energy compared to the point from which it was shot. At the highest point, the vertical velocity will be zero. We can calculate the change in height by finding the maximum height reached using the vertical component of the initial velocity.
First, let's find the vertical component of the initial velocity:
v_y = v * sin(theta)
Where:
v_y is the vertical component of the initial velocity
v is the initial velocity
theta is the angle of projection
v_y = 56.41 m/s * sin(39.51°)
v_y ≈ 34.97 m/s
Next, we can find the maximum height using the equation:
h_max = (v_y^2) / (2 * g)
h_max = (34.97 m/s)^2 / (2 * 9.8 m/s^2)
h_max ≈ 61.70 m
Finally, we can calculate the gain in potential energy using the equation:
ΔPE = m * g * Δh
Δh = h_max
ΔPE = 3.67 kg * 9.8 m/s^2 * 61.70 m
ΔPE ≈ 2,228.42 J
Therefore, the gain in potential energy at the highest point of the trajectory is approximately 2,228.42 Joules.
To find the gain in potential energy of the cannonball at the highest point of its trajectory, we need to determine the change in its height.
At the highest point of the trajectory, the vertical component of the cannonball's velocity becomes zero. We can use this fact to find the time it takes for the cannonball to reach the highest point. Let's break down the initial velocity into its horizontal and vertical components:
Vertical component of initial velocity (v₀y) = v₀ * sin(θ)
v₀y = 56.41 m/s * sin(39.51°)
v₀y = 35.93 m/s
The time taken to reach the highest point can be found using the equation for motion in the vertical direction:
Δy = v₀y * t - 0.5 * g * t²
Since the vertical displacement, Δy, at the highest point is zero, we can rewrite the equation as:
0 = v₀y * t - 0.5 * g * t²
Simplifying this equation gives us a quadratic equation:
0.5 * g * t² - v₀y * t = 0
Solving this quadratic equation for t, we get:
t = 0 (one solution)
t = 2 * v₀y / g (another solution)
Now, we can calculate the time taken to reach the highest point:
t = 2 * 35.93 m/s / 9.8 m/s²
t ≈ 7.33 s
Using this time, we can calculate the change in height, which gives us the gain in potential energy:
Δy = v₀y * t - 0.5 * g * t²
Plugging in the values:
Δy = 35.93 m/s * 7.33 s - 0.5 * 9.8 m/s² * (7.33 s)²
Δy ≈ 92.65 m
Therefore, the gain in potential energy of the cannonball at the highest point of its trajectory is approximately 92.65 m.