The maximum height the bullet will reach is 627 ft. The time(from the point when bullet was fired) it will take the bullet to hit the ground is 12.51 sec approx. The time when the bullet will be 403 ft above the ground are 2.475 sec and 10.02 sec approx.
How to obtain the maximum value of a function?
To find the maximum of a continuous and twice differentiable function f(x), we can firstly differentiate it with respect to x and equating it to 0 will give us critical points.
Putting those values of x in the second rate of function, if results in negative output, then at that point, there is maxima. If the output is positive then its minima and if its 0, then we will have to find the third derivative (if it exists) and so on.
For this case, the height of the bullet from the ground is modeled by the equation:
[tex]h(t) = -16t^2 + 200t + 3[/tex]
where height is in meters and time is in seconds. And t is the time in seconds passed from the point of time when the bullet was fired.
- A) Finding maximum height reached by the bullet:
Finding its first and second rate with respect to the variable 't', we get:
[tex]\dfrac{d(h(t))}{dt} = h'(t) = -32t + 200\\\\\dfrac{d^2(h(t))}{dt^2} = h''(t) = -32 < 0[/tex]
Thus, the second rate is negative no matter what the value of 't' is. So all the critical points would be corresponding to maxima.
Now, finding the critical points by equating the first rate = 0, we get:
[tex]h'(t) = 0\\\\-32t + 200 = 0\\\\t = \dfrac{200}{32} = 6.25 \: \rm sec.[/tex]
Due to only one critical point, and that each critical point is maxima, we have the value of h(t) globally maximum when t = 6.25
Putting t = 6.25 in the function h(t), we get the maximum height achieved by the bullet as:
[tex]h(t) = -16t^2 + 200t + 3\\h(6.5) = -16(6.5)^2 + 200(6.5) + 3 = 627 \: \rm ft[/tex]
- B) Finding the time it will take the bullet to fall on the ground
When the bullet will fall on the ground, the value of h(t) would be 0.
Putting h(t) = 0, and finding the values of 't' for which this is true, we get:
[tex]h(t) = -16t^2 + 200t + 3\\0 = -16t^2 + 200t + 3\\\\16t^2 -200t -3 =0\\\\t = \dfrac{-(-200) \pm \sqrt{ (-200)^2 - 4(16)(-3)}}{2(16)}\\\\t \approx \dfrac{200 \pm 200.479}{32}\\\\t \approx -0.0149 \: \rm sec., t \approx 12.51 \: sec.[/tex]
Bullet will fall only after it is shot, so time taken for the bullet to fall compared to the time when its shot would be greater, so time passed (t here) would be positive.
Thus, at approx t = 12.51 sec, the bullet will fall on the ground.
- C) When will the bullet be at 400 ft of height from the ground?
At t = 0, bullet was at 3 ft, and therefore, whenever bullet will reach 400 ft, the time t would be > 0.
Putting h(t) = 400, we get:
[tex]h(t) = -16t^2 + 200t + 3\\400 = -16t^2 + 200t + 3\\\\16t^2 -200t +397 =0\\\\t = \dfrac{-(-200) \pm \sqrt{ (-200)^2 - 4(16)(397)}}{2(16)}\\\\t \approx \dfrac{200 \pm 120.797}{32}\\\\t \approx 2.475\: \rm sec., t \approx 10.02\: sec.[/tex]
These both values are true. The time when t = 2.475 approx, the bullet would be going up, passing by 400 ft height, and when t = 10.02 approx, the bullet would be coming down, passing by 400 ft height.
Thus, the maximum height the bullet will reach is 627 ft. The time(from the point when bullet was fired) it will take the bullet to hit the ground is 12.51 sec approx. The time when the bullet will be 403 ft above the ground are 2.475 sec and 10.02 sec approx.
Learn more about maxima and minima of a function here:
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