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kirill115 [55]
3 years ago
6

A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far

(in meters) will the shot go if we assume they do not reach over the stop board?
Physics
1 answer:
liq [111]3 years ago
3 0

Answer:

x = 17.88[m]

Explanation:

We can find the components of the initial velocity:

(v_{x})_{o}  = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o}  = 13.3*sin(41.5)=8.81[m/s]

We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.

g = - 9.81[m/s^2]

Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.

y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]

With this time we can calculate the horizontal distance:

x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]

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The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point
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Answer:

The polar coordinate of P(x,y) = (-3.50\,m,-2.50\,m) is P (r,\theta) = (4.301\,m, 215.538^{\circ}).

Explanation:

Given a point in rectangular form, that is P(x,y) = (x,y), its polar form is defined by:

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Where:

r - Norm, measured in meters.

\theta - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

r = \sqrt{x^{2}+y^{2}} (2)

And direction is calculated by following trigonometric relation:

\theta = \tan^{-1} \frac{y}{x} (3)

If we know that x = -3.50\,m and y = -2.50\,m, then the components of coordinates in polar form is:

r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}

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Since x < 0\,m and y < 0\,m, direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula:

\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)

\theta \approx 215.538^{\circ}

The polar coordinate of P(x,y) = (-3.50\,m,-2.50\,m) is P (r,\theta) = (4.301\,m, 215.538^{\circ}).

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