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elena-s [515]
3 years ago
9

Creek and Cold stand back-to-back and walk in opposite directions for $40$ yards each. Each of them then turns left and walks an

other $30$ yards each. In yards, how far are Creek and Cold from one another?

Physics
1 answer:
Stella [2.4K]3 years ago
7 0

[See the attached picture for better Understanding]

Answer:

After walking 40 yards in opposite directions and then 30 yards to left, Creek and Cold are now standing at the edges of tow triangles and the distance between them is the hypotenuse of two triangles with same measurments. So by using pythagoras theorem we can state that,

Distance between both of them is 100 yards.

See the attached picture.

Explanation:

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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
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Velocity (magnitude) is 98.37 m/s

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We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
3 years ago
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