Well, it looks like you're asking me to prove a mathematical equation related to the motion of a lead shot. As a Clown Bot, I must admit that I'm not the best when it comes to mathematical proofs, but hey, let's give it a shot, shall we?
To start, let's break down the motion of the lead shot. We know that the time it takes to reach the maximum height, t, is given by t = U * sin(θ) / g, where U is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.
Now, let's find the maximum height reached by the lead shot, denoted as h_max. We need to analyze the shot at the exact moment when it reaches the highest point in its trajectory. At this point, the vertical component of its velocity is zero.
Using the equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can say that at the highest point, the final velocity v is 0, the initial velocity u is U * sin(θ), and the acceleration a is -g (since gravity pulls objects downwards).
Therefore, we can write the equation as:
0 = U * sin(θ) - g * t_max
Solving for t_max, we have:
t_max = U * sin(θ) / g
Now, we can substitute this value of t_max in the equation for the height h_max using the equation of motion, s = ut + 0.5at^2, where s is the displacement:
h_max = (U * sin(θ) * t_max) - (0.5 * g * t_max^2)
Substituting the value of t_max, we get:
h_max = (U * sin(θ) * (U * sin(θ) / g)) - (0.5 * g * (U * sin(θ) / g)^2)
Simplifying further:
h_max = (U^2 * sin^2(θ)) / g - (0.5 * g * (U^2 * sin^2(θ) / g^2))
Canceling out common terms:
h_max = (U^2 * sin^2(θ)) / g - (0.5 * U^2 * sin^2(θ) / g)
Adding the fractions:
h_max = (U^2 * sin^2(θ) - 0.5 * U^2 * sin^2(θ)) / g
Simplifying:
h_max = (0.5 * U^2 * sin^2(θ)) / g
Finally, we arrive at the expression:
h_max = U^2 * sin^2(θ) / (2 * g)
Voila! There you have it, the proof that the greatest height reached by the lead shot is indeed h_max = U^2 * sin^2(θ) / (2 * g). I hope I was able to bring some mathematical clownery to your day!