When will the ball reach the water below the cliff?

Question

A ball is thrown into water from a cliff that is 175 m high. The height of the ball above the water after it is thrown is modelled by the function h(t)= -5t^2 + 10t + 175, where h(t) is the height in metres and t is time in seconds.

Answer 1

Victoria,

    Just think of the water as the X-axis.  The 175 meter cliff is the Y-axis, where t=0.  Once the ball is thrown, time is

 

moving forward. 

 

 Question:  When will the ball hit the water?  It will hit the water when the height is zero.

 

 

Let h(X)=0=y                  0=-5t² +10t +175      Simplify the equation by dividing every term by -5.

 

                                     0= t² -2t-35                Factor

 

                                     0=(t+__)(t-__)     One answer will be positive, and one will be negative.    

 

Don't use any negative time in your answer because time is not negative.

 

 

You can also enter this equation into the graphing calculator by changing t=X .  Also, the window would need to be

 

changed to view the parabola better.  Here is what I suggest, and it should give you a good view.  Press the 2nd

 

function button, WINDOW.  Then change your information to look like the data below:

 

                                                    Let X min=0

                                                         X max=10

                                                         y min= -1

                                                        y max= 200

 

 

I hope that I have helped you.  If I have, please give me a thumbs up.

 

                                                         Susan C.