[Video] Q34: Y = -16t^2 + 64t + 80the equation above gives the height of an object above the ground, y, in feet, t seconds after it is launched from a platform. how many seconds after it is launched does the object reach the ground?

Explanation for Question 34 From the Math (Calc) Section on the 2019 March Sat

Question 34 says the equation above gets the height of an object above the 2 ground. Y in feet, T seconds, 3 after it is launched from a platform, how many seconds after this launched, 4 does the object reach the ground since Y represents the height in this case, 5 we're going to set Y equal to zero. 6 So we can find when, what the, 7 or how many seconds it will take for the object to be zero feet 8 off the ground. Now, 9 once we do that, we're going to have to factor. 10 So we're going to divide both sides of the equation by negative 16. 11 And then we get T squared plus negative 40 plus 12 negative five. Now this factors into T minus five times 13 T plus one. And so we know that T is 14 equal to five and T is equal to negative one. 15 Now you can't have a negative amount of seconds. 16 Time can't be negative. So the only answer here that makes sense is T 17 is equal to five. And so we know that the answer has to be 18 five.

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