An arrow is shot vertically upward at a rate of 220ft/s. Use the projectile formula h=?16t2+v0t to determine at what time(s), in seconds, the arrow is at a height of 400ft. Round your answer(s) to the nearest tenth of a second.
Accepted Solution
A:
Answer:t1 = 2.2 st2 = 11.5 sStep-by-step explanation:The general quadratic formula for these cases ish(t) = h0 + v*t + 0.5*a*t^2 Where a = the acceleration of the body. In this case, since the arrow is going upwards, we take the acceleration of the gravity.a = -32.2 feet/s^2 = -9.8 m/s^2v = the velocityv = 220 ft/sh0 = the initial distance. We assume it launched from groundh0 = 0h(t) = - 16.1*t^2 + 220*tSee, attached pictureWhen the height h = 400 ft - 16.1*t^2 + 220*t = 400t1 = 2.159 s ≈ 2.2 st2 = 11.505 s ≈ 11.5 s