Question

the height of a projectile, in feet, t seconds after launch is modeled by the equation... how long does the projectile return to the ground? options: 2 seconds, 4 seconds, 14 seconds, 16 seconds

Explanation:

Step1: Analyze the problem (free fall motion)

The problem is about a projectile (probably a ball) thrown upward, and we need to find the time to return to the ground. For free - fall (assuming no air resistance), the time of flight (time to go up and come back down) can be related to the initial vertical velocity. But if we assume the initial upward velocity \(v_0\) and the acceleration due to gravity \(g = 9.8\ m/s^2\) (acting downward). The time to reach the maximum height \(t_{up}\) is given by \(v = v_0+at\), at maximum height \(v = 0\), so \(t_{up}=\frac{v_0}{g}\). The time to come back down from the maximum height to the ground is the same as the time to go up (because the motion is symmetric in the absence of air resistance), so the total time of flight \(T=\frac{2v_0}{g}\). But maybe in this problem, it's a simpler case. If we assume that the time to go up is \(t\) and the time to come down is also \(t\), and maybe the initial time given is the time to go up. Wait, the options are 2, 4, 14, 16 seconds. Wait, maybe it's a case where the time to go up is 2 seconds, then the time to come down is also 2 seconds, so total time is \(2 + 2=4\) seconds? Wait, no, maybe the initial upward motion takes \(t\) seconds, and the downward motion takes the same \(t\) seconds. Wait, maybe the problem is about a ball thrown upward with an initial velocity, and the time to reach the top is \(t\), then the time to come back is \(t\), so total time is \(2t\). If the time to go up is 2 seconds, total time is 4 seconds? Wait, maybe the original problem (even though the image is a bit unclear) is about a projectile where the time to ascend is 2 seconds, so the time to descend is also 2 seconds, so total time of flight is \(2\times2 = 4\) seconds? Or maybe it's a different scenario. Wait, let's think again. In free - fall, the time of flight for a projectile thrown vertically upward with initial velocity \(v_0\) is \(T=\frac{2v_0}{g}\). If we consider that at the top, \(v = 0\), and \(v=v_0 - gt_{up}\), so \(t_{up}=\frac{v_0}{g}\). Then \(T = 2t_{up}\). If \(t_{up}=2\) seconds, then \(T = 4\) seconds. So the answer should be 4 seconds.

Answer:

4 seconds (corresponding to the option "4 seconds")