A projectile is fired straight up from ground level. After t seconds its height s, in feet above the ground, is given by s=350t-16t^2.

Question

A projectile is fired straight up from ground level. After t seconds its height s, in feet above the ground, is given by s=350t-16t^2. During what time interval is the projectile more than 1250 ft above the ground?

Answer 1

So we should probably first find out if the projectile reaches a height of 1250 ft. at all. Let's look at what time(s) the projectile is at a height of 1250 ft. . To do this, lets assume that the projectile reaches the appropriate height at some time t_f.

s(t_f) = 350t_f - 16(t_f)² = 1250

Let's factor the polynomial so that we can solve for t_f

350t_f - 16(t_f)² = 1250

-1250 -1250

-16(t_f)² + 350t_f - 1250 = 0

(-1/16)(-16(t_f)² + 350t_f - 1250) = (-1/16)*0

(t_f)² - 21.875t_f + 78.125 = 0

To solve for t_f, it seems we will need to make use of the quadratic formula

t_f = (-b ± √(b² - 4ac)) / (2a)

For this polynomial, a = 1, b = 21.875, c = 78.125. Plugging these values into the formula we get that t_f = 4.495 and 17.38 seconds. This means that the projectile passes through the height of 1250 feet as it goes up at 4.495 seconds, and once again as it falls back down at 17.38 seconds clearly the interval is (4.495, 17.38) seconds, or 4.495 < t < 17.38 seconds. To check, just graph the projectile motion formula given in the problem in to a graphing calculator and visually inspect when the height of the projectile is greater that 1250 feet. Make sure you adjust the answer to reflect the correct number of significant figures given in the this problem.