An object has been thrown straight up into the air. the formula h = vt - 16t^2 gives the height of the object above the ground after t seconds, when it is thrown upward with an initial velocity v. after how many seconds will the object hit the ground if it is thrown with a velocity of 128 feet per second?

8 seconds.

Step-by-step explanation:

To solve this question, we need to find the time, , when , that is, when the height of the object above the ground is 0 feet.

Substituting feet per second and gives  the quadratic:

Since 128 is divisible by 16, it can be reduced to .

We must now solve for .

We can easily see that one answer to the equation is 0,

(we need not concern ourselves with the 16 outside of the parenthesis as in the equation above, since 16 multiplied by 0 is 0). However this is the time the object is released into the air.

The second answer, is also easy to see by inspection:

.

Therefore the object lands 8 seconds after it is thrown.

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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step-by-step explanation: