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3 years ago

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Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche

d at the same initial velocity for two different launch angles. Discuss the case where the ranges are the same.

1 answer:

DaniilM [7]3 years ago

5 0

Answer:

$R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}$

Explanation:

Using kinematics equations:

$\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t$

Use $\Delta y = 0$ due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

$t=\frac{2v_{0y}}{g}$

Use the expression in the first equation to have

$R = \frac{2v^2 \cos\theta\sin\theta}{g}$

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

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Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

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To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

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R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

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Our values are

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Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

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At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

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Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

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3 years ago