Quadratic Equations in Word Problems

Question

The height, h meters, of a ball projected directly upwards from the ground can be modeled by h = 56t - 7t^2, where t is the time in seconds after it leaves the ground.

 

(i) Find the height of the ball 3.5 seconds after it leaves the ground.

(ii) At what time will the ball strike the ground again?

(iii) When will the ball be 49m above the ground? Briefly explain why there are two possible answers.

Answer 1

(i) The height is a function of time, in seconds. So here we can just plug-in t=3.5 and solve for h. So we get h=56(3.5)-7(3.5)² =110.25. So the height is 110.25 meters.

 

(ii) The ball will strike the ground when the height is zero. Thus, we want to set h=0 and solve for t.

 

               0=56t-7t²

               0=t(56-7t)                    we pull out the gcf of t to factor

               0=t         0=56-7t         we set each factor equal to zero and solve

               0=t         7t=56

               0=t         t=8

So the ball hits the ground again at 8 seconds.

 

(iii) The ball follows a parabolic path and will thus hit a height of 49 meters twice. Once when it is rising and once when it is falling. We must set h=49 and solve for t.

 

               49=56t-7t²

               7t²-56t+49=0             set the equation to 0

               7(t²-8t+7)=0              to make factoring simple we can pull out a 7

               t²-8t+7=0                  divide both sides by 7

               (t-7)(t-1)=0

               t-7=0      t-1=0

               t=7         t=1

So the ball will be 49 meters above the ground at 1 second and at 7 seconds.