Answer:
After 3 seconds while going up and 6 seconds while coming down
Step-by-step explanation:
Since projectile is launched from ground level with an initial velocity of v 0 feet per second, initial height =0
Height is given by
where v0 = initial velocity = 144
a) When h =288 ft.
At t= 3 or 6 seconds the projectile would be at height 288 ft.
So while going up after 3 seconds it wouldbe at a height of 288 ft and while coming after 6 seconds.
9514 1404 393
Answer:
13 in by 51 in
Step-by-step explanation:
The area is the product of the dimensions, so is ...
663 = x(4x -1)
4x^2 -x -663 = 0 . . . . . . subtract 663 to put in standard form
Using the quadratic formula, we can find the solutions.
x = (-(-1) ±√((-1)^2 -4(4)(-663)))/(2(4))
x = (1 ± √10609)/8 = (1 ±103)/8
Only the positive solution is of any use in this problem, so ...
x = 104/8 = 13
4x-1 = 4(13)-1 = 51
The dimensions of the rectangle are 13 inches by 51 inches.
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I find it easiest to solve these using a graphing calculator.
