QuestionA shot reaches a max height of 21.3 ft at 40 ft from release. Find its max horizontal distance using $F(x)=-0.01x^2+0.8x+5.3$ .

Studdy Solution

STEP 1

Assumptions1. The maximum height of the shot is21.3 feet. The maximum height occurs40 feet from the point of release3. The equation for the trajectory of the shot is given as $(x)=-0.01x^+0.8x+5.3$
4. We are to find the initial velocity and the angle of projection to calculate the maximum horizontal distance5. We assume the shot is projected from the ground level6. We assume there is no air resistance

STEP 2

a. The maximum height of the shot and how far from its point of release does this occur?
The maximum height of the shot is given as21. feet and it occurs40 feet from the point of release.

STEP 3

b. To find the shot's maximum horizontal distance, we first need to find the initial velocity and the angle of projection.
The trajectory of the shot is given by the equation $(x)=-0.01x^2+0.8x+5.3$ . This is a quadratic equation in the form of $y = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are constants.

STEP 4

The initial velocity $v0$ can be calculated using the formula for the maximum height $H$ of a projectile, which is given by $H = \frac{v0^2 \sin^2(\theta)}{2g}$ , where $\theta$ is the angle of projection and $g$ is the acceleration due to gravity.
Rearranging the formula to solve for $v0$ , we get $v0 = \sqrt{\frac{2gH}{\sin^2(\theta)}}$

STEP 5

We know that the maximum height $H$ is21.3 feet and the acceleration due to gravity $g$ is approximately32.2 feet/second^2. However, we don't know the angle of projection $\theta$ .We can find $\theta$ by using the fact that the maximum height occurs40 feet from the point of release. At this point, the derivative of the trajectory function $'(x)$ is zero.
So, we first find the derivative of $(x)$ $'(x) = -0.02x +0.8$

STEP 6

Setting $'(x)$ equal to zero and solving for $x$ gives us the x-coordinate of the maximum height $-0.02x +0.8 =0$ olving this equation gives $x = \frac{0.8}{0.02} =40\, feet$

STEP 7

At $x =40$ feet, the y-coordinate is the maximum height, which is21.3 feet.So, the coordinates of the maximum height are $(40,21.3)$ .

STEP 8

The tangent of the angle of projection $\theta$ at the maximum height is given by the derivative of the trajectory function $'(x)$ at $x =40$ feet.So, we have $\tan(\theta) = F'(40)$

STEP 9

Substituting $x =40$ feet into $'(x)$ gives $\tan(\theta) = -.02(40) +.8 =$ So, $\theta = \arctan() =$ degrees.

STEP 10

Substituting $H =21.3$ feet, $g =32.2$ feet/second^2, and $\theta =0$ degrees into the formula for $v0$ gives $v0 = \sqrt{\frac{2 \times32.2 \times21.3}{\sin^2(0)}}$ However, since $\sin(0) =0$ , this formula is undefined.This suggests that the shot was thrown straight up, which is consistent with the fact that the angle of projection is0 degrees.

STEP 11

The maximum horizontal distance of the shot is the same as the distance from the point of release to the maximum height, which is40 feet.
So, the shot's maximum horizontal distance is40 feet.

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