Solve the problem. Assume the x-axis is horizontal, the positive y-axis is vertical (opposite g), the ground is horizontal, and only the gravitational force acts on the objects
-A projectile is fired with an initial speed of 585 m/sec at an angle of 45°. What is the greatest height reached by the projectile? Round your answer to the nearest tenth.
A) 85,556.3 m
B) 8730.2 m
C) 34,920.9 m
D) 84.4 m

Question

Quizplus · Accepted Answer

The greatest height $H$ reached by a projectile launched with an initial speed $v_0$ at an angle $\theta$ to the horizontal is given by the formula $H = \frac{v_0^2 \sin^2(\theta)}{2g}$, where $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$). For $v_0 = 585 \, \text{m/s}$ and $\theta = 45^\circ$, $H = \frac{(585)^2 \sin^2(45^\circ)}{2 \times 9.8}$ which calculates to approximately $8730.2 \, \text{m}$.

Solve the Problem