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4 votes
A toy rocket is launched vertically upward from a 12 foot platform with an

initial velocity of 128 feet per second. Its height h at timet seconds after
launch is given by the equation h(t) = -16t2 + 128t + 12. How many
seconds until the rocket is 40 feet?
0.23 seconds
4.5 seconds
5.67 seconds
7.77 seconds
8.09 seconds
11.34 seconds
12.52 seconds

User DMrFrost
by
8.4k points

1 Answer

2 votes

Answer:

The rocket is at 40 feet after 7.77 seconds.

Explanation:


h(t) = -16t^(2) + 128t + 12\\40= -16t^(2) + 128t + 12\\16t^(2) - 128t + 28 = 0\\4(4t^(2) - 32t + 7) = 0\\

We use the quadratic formula to solve for t.


t=\frac{32+\sqrt{32^(2)-4*4*7} }{2*4} \\t=(32+√(1024-112) )/(8)\\t=(32+√(912) )/(8)\\t=4+(√(57) )/(2)\\t=\frac{32-\sqrt{32^(2)-4*4*7} }{2*4} \\t=(32-√(1024-112) )/(8)\\t=(32-√(912) )/(8)\\t=4-(√(57) )/(2)

We take the positive answer which is 7.77

User Aymand Osanes
by
9.1k points
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