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The height, in feet, after x seconds of an object launched straight up can be found by the function h(x)=−16x2+v0x+h0, where v0 is the initial velocity of the object and h0 is the initial height.

A ball is launched straight up into the air from a height of 16 ft with an initial velocity of 60 ft/s.

After how many seconds does the ball hit the ground?



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Luv2Teach
if you knew calculus this problem would take like 2 seconds, but since we are working with maybe Algebra 2, you have to use the equation as it is and find where h(x) is 0, because that is the height of something when it hits the ground. Meaning, you have to factor this and solve it for x.  Here's your equation:
h(x) = -16x^2 + 60x +16.  Use the quadratic formula to find that the values for x are -.25 and 4 seconds.  Of course time cannot ever carry a negative value, so the answer is 4 seconds.  It takes the ball 4 seconds to hit the ground.

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