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Sagot :
A cannonball's height, h, is given by the equation:
[tex]\text{ h = 160t - 16t}^2[/tex]Let's determine the maximum height.
Let's get the first derivative of the equation.
[tex]\text{ h = 160t - 16t}^2[/tex][tex]\frac{\text{ dh}}{\text{ dt}}\text{ = 160}(1)\text{ - 16}(\text{2})\text{t = 160 - 32t}[/tex]dh/dt represents the speed of the cannonball.
At maximum height, the ball no longer goes up, thus, the speed (dh/dt) is 0.
Let's now determine the time (t) when the cannonball reaches the maximum height.
[tex]\frac{\text{ dh}}{\text{ dt}}\text{ = 160 - 32t}\frac{}{}[/tex][tex]\text{ 0 = 160 - 32t}[/tex][tex]\text{ 32t = 160}[/tex][tex]\frac{\text{ 32t = 160}}{\text{ 32}}[/tex][tex]\text{ t = 5 seconds}[/tex]Therefore, in 5 seconds, the cannonball reaches its maximum height after being shot.
Let's now determine the maximum height, at t = 5 seconds.
[tex]\text{ h = 160t - 16t}^2[/tex][tex]\text{ h = 160}(\text{5})\text{ - 16}(\text{5})^2\text{ = 160}(5)\text{ - 16}(\text{25})[/tex][tex]\text{ h = 800 - 400 = 400 meters}[/tex]Therefore, the maximum height of the cannonball after being shot is 400 meters.
The answer is Choice C: 400 meters.
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