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A projectile is launched from a catapult with initial velocity 350 feet/sec at an angle of 65°.
-Find the time(s) required to reach a height of 200 feet? Round to the nearest hundredth.
A) approximately 0.65 sec.
B) approximately 20.30 sec.
C) approximately 0.65 sec. and 19.17 sec.
D) approximately 0.65 sec. and 20.30 sec.

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Quizplus · Accepted Answer

The time(s) to reach a height of 200 feet can be found by using the vertical component of the initial velocity and the formula for vertical motion under gravity. The vertical component of the velocity is $v_{0y} = v_0 \sin(\theta)$, where $v_0 = 350$ feet/sec and $\theta = 65^\circ$. The equation for the height $h$ at time $t$ is $h = v_{0y}t - \frac{1}{2}gt^2$, where $g$ is the acceleration due to gravity (approximately $32$ feet/sec$^2$ downward). Solving this quadratic equation for $t$ when $h = 200$ feet gives two possible times, one during the ascent and one during the descent of the projectile. The correct times, rounded to the nearest hundredth, are approximately 0.65 sec and 19.17 sec, which matches option C.

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