Question

(Projectile Motion) Mary is outside her apartment building. She needs to pass her key to John, who is 6 m above the ground floor of the building. Mary has no time to run upstairs, so she throws the key towards John. Suppose the height, h m, of the key above the ground floor at time t seconds is given by h = -11t² + 20t + 1. Is it possible for John to catch the key? Justify your answer.​

Answer 1

To determine if it is possible for John to catch the key, we need to find the time when the height of the key is equal to 6m, which is John's height above the ground floor. We can set the equation h = 6 and solve for t:

-11t^2 + 20t + 1 = 6
-11t^2 + 20t - 5 = 0

Now we can use the quadratic formula to solve for t:

t = (-b ± √(b^2 - 4ac)) / 2a
where a = -11, b = 20, and c = -5

t = (20 ± √(20^2 - 4(-11)(-5))) / 2(-11)
t = (20 ± √(400 + 220)) / -22
t = (20 ± √620) / -22
t = (20 ± 2√155) / -22
t = (20 ± 2√3 * 5^2) / -22
t = (20 ± 2 * 15) / -22
t = (20 ± 30) / -22

t = (-10 ± 30) / -22
t = (-10 + 30) / -22 = 20 / -22 = -15 / 11
t = (-10 - 30) / -22 = -40 / -22 = 20 / 11

There are two solutions for t: -15 / 11 and 20 / 11. However, the time can not be negative as it represents the amount of time that has passed. Therefore, the only valid solution is t = 20 / 11.

So, the key reaches John's height at t = 20 / 11 seconds, which means it is possible for John to catch the key