A projectile is thrown upward so that it's distance above the ground after t seconds is H(x)=-12x^2+312x.

Question

(a). After how many seconds does it reach it's maximum height?

(b). After how many seconds does it hit the ground?

Answer 1

The question should state that the time is represented by x, not t.

H(x)=-12(x²-26x)=-12(x²-26x+169-169)=-12((x-13)²-169)=12(169-(x-13)²).

a) The smallest value of (x-13)² is zero, when x=13, making the value in brackets maximum. So x is the time it reaches its max height of 12×169=2028 feet. Therefore the maximum height is reached in 13 seconds.

b) When H(x)=0, x(x-26)=0 and x=0 (start) or 26 seconds. So it takes 26 seconds to fall to the ground.