Use s = 96 + 80t - 16t^2 to answer the question given below.

QUESTION:

After how many second t will the ball pass the top of the building on its way down?

My work:

Let s = 96 feet...Is this right?

I then equated 16t^2 -80t = 96

Is this correct?

If so, what is my next step?

Thanks

Gilligan,

The expression s = 96 + 80t - 16t^{2} tells you the distance the ball is above the ground in feet at t seconds after the the ball is thrown upwards. Notice that when t = 0 seconds the expression gives s = 96 feet. This means that when the ball was released it was 96 feet above the ground, at the top of the building. The ball proceeds upwards and then downward and the question asks at what time the ball will be at a height of 96 feet again. Thus, as you said you let s = 96 and then the equation becomes

96 = 96 + 80t - 16t^{2}

Solve this equation for t. You will find two solutions, one being t = 0.

Penny

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