The height of a projectile launched at an angle of $\theta$ degrees from the ground with an initial velocity $v_{0}$ from a starting height of $s$ can be found using the formula $y=-16 t^{2}+v_{0} \cdot \sin (\theta) t+s$. A projectile is launched from a height of 25 feet at an angle of $30^{\circ}$ from the ground with an initial velocity of $140 \mathrm{ft} / \mathrm{s}$. Find the maximum height that the projectile reaches, using the fact that $\sin \left(30^{\circ}\right)=0.5$. When computing maximum height in the following problem, round both the time and the computed height to the nearest tenth.
A) 101.6 feet
B) 125.1 feet
C) 118.4 feet
D) 99.3 feet

Question

Quizplus · Accepted Answer

The maximum height can be found by first determining the time at which the projectile reaches its peak. This is done by finding the derivative of the height function and setting it to zero to solve for $t$. The given formula is $y=-16t^2 + (v_0 \cdot \sin(\theta))t + s$, where $v_0 = 140$ ft/s, $\sin(\theta) = 0.5$, and $s = 25$ feet. Substituting these values gives $y = -16t^2 + (140 \cdot 0.5)t + 25$. The derivative of $y$ with respect to $t$ gives the velocity as a function of time, $v(t) = -32t + 70$. Setting $v(t) = 0$ to find the time at peak height gives $t = 2.1875$ seconds. Substituting $t = 2.2$ seconds (rounded to the nearest tenth) back into the original equation gives the maximum height, which is approximately $101.6$ feet when rounded to the nearest tenth.

The Height of a Projectile Launched at an Angle Of \theta