a model rocket is fired up into the air. the height, h, in meter, above the ground at any time, t, in secounds, after the engine has shut down is given by h=-5t^2 + 50t + 195. Determine the time that elapses before the rocket hits the ground.

Question

a_clan · Answer

put h=0 i.e the height of  ground

a_clan · Answer

you will  get two values of time, one indicating the time before it took off and the other indicating the time when it will touch the ground again.

anonymous · Answer

set h to zero and solve
$$-5t ^{2}+50t+195 = 0$$
$$5(-t ^{2}+10t+39)=0$$
(t + 3)(-t + 13) = 0
solving for t we get t = -3 and t = 13, throw away the -3 since it can't hit the ground before it takes off, and the answer is 13 seconds.

a_clan · Answer

t=-3 indicates that the rocket did not took off from ground but from a certain height.

anonymous · Answer

Actually, the equation was for the time *after* the engine went out, so we can assume that the engine burned for 3 seconds to propel the rocket to its max height. Does that make sense?

a_clan · Answer

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