Assigment 10

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course Mth 277

7/11 midnight

Question: Find the time of flight Tf (to the nearest tenth of a second) and the range Rf (to the nearest unit) of a projectile fired (in a vaccum) from ground level at `alpha = 65.54 degrees and v0 = 19.07 m/s. Assume that g = 9.8 m/s^2.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If the v0 = 19.07 m/s, then use the Pythagorean theorem:

vy0 = sin alpha * v0 = 17.4 m/s

vx0 = cos alpha * v0 = 7.9 m/s

The time of flight equals vy0/g, where g is acceleration due to gravity.

Tf = vy0/g = 17.4/9.8 = 1.78 seconds

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It appears you've relied on your knowledge of physics to solve this problem.

Setting this up in terms of a vector function:

The acceleration function is

`a(t) = 0 `i -g `j.

Integrating once with respect to time yields the velocity function

`v(t) = c_1 `i + (c_2 - g t) `j.

If `v(0) = vx_0 `i + vy_0 `j then c_1 = vx_0 and c_2 = vy_0, so our velocity function is

`v(t) = vx_0 `i + (vy_0 - g t) `j.

The position function is found by integrating the velocity function. If the initial position is taken to be 0 `i + 0 `j, then we have

`r(t) = vx_0 t `i + (vy_0 t - 1/2 g t^2) `j

The maximum height would be achieved when `v(t) is horizontal, i.e., when its `j component is zero:

vy_0 - g t = 0

which occurs when t = vy_0 / g.

The horizontal range is the change in horizontal position between firing and returning to the ground.

The return to the ground occurs when the `j component of the position function is zero:

vy_0 t - 1/2 g t^2 = 0

which has solutions t = 0 and t = 2 vy_0 / g.

When t = 2 vy_0 / g the position function is

`r(2 vy_0 / g)

= vx_0 * vy_0 / g `i + (vy_0 * 2 vy_0 / g - 1/2 g (2 vy_0 / g)^2 )

= 2 vx_0 vy_0 / g `i + 0 `j.

So the horizontal range is 2 vx_0 vy_0 / g.

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Question: An object is moving along the curve r = 1/(1 - sin(theta)), theta = t - pi/2. Find its velocity and acceleration in terms of the unit polar vectors u_r and u_theta.

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Your solution:

Position = 1/ (1 - sin(t - pi/2))

Velocity = sin(t)/( cos(t) + 1)^2

Acceleration = -0.25*(cos t - 2) * sec^4(t/2)

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Position, velocity and acceleration are all vector quantities.

You've taken the derivatives of r with respect to t, but this does not constitute the velocity and accleration functions.

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Given Solution:

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Question: If a shotputter throws a shot from a height of 5.5t and an angle of 53 degrees with initial speed 28 ft/s. What is the horizontal distance of the throw?

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Your solution:

v0 = 28 ft/s

vy0 = sin theta * v0 = 22.4 ft/s

vx0 = cos theta * v0 = 16.9 ft/s

flight time = vy0/g = 2.29 seconds

distance y = vy0 * flight time = 51.3 feet

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Problems in this section should be set up in terms of vector functions.

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Given Solution:

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Question: A child running along level ground at the top of a 40ft high vertical cliff at a speed of 15ft/s, throws a rock over the cliff into the sea below. If the rock is released 10 ft from the edge and at an angle of 45degrees, how long does it take the rock to hit the water and how far away from the base of the cliff does it hit?

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Your solution:

v0 = 15 ft/s

vx0 = 10.6 ft/s

vy0 = 10.6 ft/s

*time to apex = 1.08

*flight time = 3.05 seconds

x distance = vx0 * flight time = 32.4 feets

Confidence rating: 3

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Given Solution:

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Question: **A gun is fired with muzzle speed 700ft/s at an angle of 20degrees. It overshoots the target by 60 ft. If the target is moving away from the gun at a constant speed of 15ft/s and the gunner takes 30 seconds to reload, at what angle should the second shot be fired with the same muzzle speed?

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Your solution:

vx0 = cos20*700 = 658 ft/s

vy0 = sin20*700 = 239 ft/s

time of impact = 134.194 seconds

x = 32140 ft

t2 = 30.09125 s

x2 = 471.7679 feet (when the gun fires again)

When the second shot fires, we want to find theta so that the bullet collides with the target, which means that the net y change is 0.

Therefore, the vy1*t should equal 0.5*g*(t^2).

Using this equation, we find that the initial velocity in the y direction should be 659 feet/s.

Using a similar equation (without deceleration), we find that the initial velocity in the x direction should be 657 feet/s.

Since our muzzle velocity must be 700 feet/s, we can use the formula:

sin(theta)*700 = vy1 = 659

theta = asin(659/700) approx = 70.29 degrees

Therefore, the gun should be aimed at 70.29 degrees above the horizontal in order to hit the target.

confidence rating #$&*:

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Given Solution:

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Most of your solutions are good, but your problems are not set up in terms of vector functions.

The common characteristic of all these problems is that

`a = -g `j

From this integration yields

`v = vx_0 `i + (vy_0 - g t) `j

and another integration yields

`r = (vx_0 t + x_0) `i + (y_0 + vy_0 t - 1/2 g t^2) `j.

I suggest you rework these problem within this framework. I don't think you'll have trouble doing so, and if you can do so with confidence there's no need to submit them. However if you have questions or want me to check your work I'll be glad to.

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