SOLUTION: A model rocket is launched with an initial velocity of 180 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 180t.
How many se
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h = −16t2 + 180t.
How many se
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Question 1073930: A model rocket is launched with an initial velocity of 180 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 180t.
How many seconds after launch will the rocket be 280 ft above the ground? Round to the nearest hundredth of a second.
.......... (smaller value)
You can put this solution on YOUR website! By your equation;
-16tē+180t=280
16tē-180t+280=0
4tē-45t+70=0
Using the quadratic formula, we get two roots of 9.38540223912 and 1.86459776088, or 9.4 and 1.9 seconds after launch. ☺☺☺☺
