Solve the problem.
-A projectile is fired with an initial velocity of 600 feet per second at an angle of 45° with the horizontal. To the nearest foot, find the maximum altitude of the projectile.
A) 2804 ft
B) 2812 ft
C) 2828 ft
D) 2795 ft

Question

Quizplus · Accepted Answer

The maximum altitude $ h $ of a projectile fired at an angle $ \theta $ with an initial velocity $ v_0 $ is given by the formula $ h = \frac{v_0^2 \sin^2(\theta)}{2g} $, where $ g $ is the acceleration due to gravity (approximately $ 32 $ feet per second squared for Earth). Given $ v_0 = 600 $ feet per second and $ \theta = 45^\circ $, and using $ \sin(45^\circ) = \frac{\sqrt{2}}{2} $, we have:$$ h = \frac{(600)^2 \left(\frac{\sqrt{2}}{2}\right)^2}{2 \cdot 32} = \frac{360000 \cdot \frac{1}{2}}{64} = \frac{180000}{64} = 2812.5 $$Rounding to the nearest foot gives $ 2812 $ feet.

Solve the Problem. -A Projectile Is Fired with an Initial Velocity of 600