Solve the problem.
-A projectile is fired with an initial velocity of 500 feet per second at an angle of $$70 ^ { \circ }$$ with the horizontal. To the nearest foot, find the maximum altitude of the projectile.
A) 3432 ft
B) 3449 ft
C) 3441 ft
D) 3465 ft

Question

Quizplus · Accepted Answer

The maximum altitude $ h $ of a projectile fired at an angle $ \theta $ with the horizontal and 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). Plugging in the given values: $ v_0 = 500 $ feet per second, $ \theta = 70^\circ $, and $ g = 32 $ feet per second squared, we get:$$ h = \frac{(500)^2 \sin^2(70^\circ)}{2 \cdot 32} \approx 3449 \text{ feet} $$Therefore, the maximum altitude of the projectile, to the nearest foot, is approximately 3449 feet.

Solve the Problem 70∘70 ^ { \circ }70∘ With the Horizontal

Solve the Problem $70 ^ { \circ }$ With the Horizontal