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

Multiple Choice

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