A cannonball of mass 6.09 kg is shot from a cannon at an angle of 47.18° relative to the horizontal and with an initial speed of 48.90 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?
h =vₒ²sin²α/2g
ΔPE= m•g•h = m•g •vₒ²sin²α/2g
To find the gain in potential energy of the cannonball at the highest point of its trajectory, we need to first find its final height at that point.
The cannonball reaches its highest point when its vertical velocity becomes zero. At this point, all its initial kinetic energy is converted into potential energy.
We can find the final height by using the vertical motion equations. Let's use the following conventions:
- Upward direction is positive.
- Acceleration due to gravity is denoted by "g" and is equal to 9.8 m/s^2.
Since the initial vertical velocity is not given, we can find it using the given initial speed and angle of projection.
Vertical initial velocity = initial speed * sine(angle of projection)
= 48.90 m/s * sin(47.18°)
= 48.90 * 0.726
≈ 35.5008 m/s
Now, we can find the time taken to reach the highest point using the following formula:
Vertical displacement = (vertical initial velocity) * time + (0.5) * (acceleration due to gravity) * (time^2)
0 = 35.5008 * t - 0.5 * 9.8 * (t^2)
Solving this quadratic equation, we get:
9.8t^2 - 35.5008t = 0
Factoring out a "t", we have:
t(9.8t - 35.5008) = 0
So, either t = 0 or 9.8t - 35.5008 = 0
Ignoring the t = 0 solution, let's solve for t:
9.8t - 35.5008 = 0
9.8t = 35.5008
t = 35.5008 / 9.8
t ≈ 3.63 s
Now that we have the time, we can find the maximum height using the following formula:
Vertical displacement = (vertical initial velocity) * time + (0.5) * (acceleration due to gravity) * (time^2)
H = 35.5008 * 3.63 - 0.5 * 9.8 * (3.63^2)
H ≈ 64.4 m
The gain in potential energy relative to the point from which the cannonball was shot is equal to the change in height.
So, the gain in potential energy is 64.4 meters.