Let's determine the formula for the maximum height [tex]\(H\)[/tex] of a projectile.
When a projectile is launched with an initial velocity [tex]\(U\)[/tex] at an angle [tex]\(\theta\)[/tex] to the horizontal, its vertical component of velocity is [tex]\(U \sin \theta\)[/tex].
1. Initial conditions:
The vertical component of the velocity at launch is given by:
[tex]\[
U_y = U \sin \theta
\][/tex]
2. Maximum height conditions:
At the maximum height, the vertical component of the projectile’s velocity will be zero because gravity will decelerate the projectile until it stops rising and starts to fall down.
3. Using kinematic equations:
The kinematic equation for vertical motion under constant acceleration due to gravity [tex]\(g\)[/tex] is:
[tex]\[
v_y^2 = U_y^2 - 2gH
\][/tex]
At maximum height, the final vertical velocity [tex]\(v_y = 0\)[/tex], hence:
[tex]\[
0 = (U \sin \theta)^2 - 2gH
\][/tex]
4. Solving for [tex]\(H\)[/tex]:
Rearrange the equation to solve for the maximum height [tex]\(H\)[/tex]:
[tex]\[
(U \sin \theta)^2 = 2gH
\][/tex]
[tex]\[
H = \frac{(U \sin \theta)^2}{2g}
\][/tex]
Simplifying the expression gives:
[tex]\[
H = \frac{U^2 \sin^2 \theta}{2g}
\][/tex]
By examining all the options given, we identify the formula for the maximum height corresponds to:
B: [tex]\(\frac{U^2 \sin^2 \theta}{g}\)[/tex]
Thus, the correct answer is:
[tex]\[
\text{Option B: } \frac{U^2 \sin^2 \theta}{g}
\][/tex]
