A ball is thrown horizontally from a 90 m cliff and strikes the ground 70 m from the base, what is the initial velocity?
1 answer:
Answer:
The initial velocity is 16 m/s
Explanation:
The position of the ball at a time t is given by the following vector:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
Where:
r = position vector at time t
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)
When the ball reaches the ground its position vector will be "r final" (see attached figure). The x-component of this vector is 70 m and the y-component is -90 m. Notice that the origin of the frame of reference is located at the throwing point (y0 and x0 = 0). Then, we can use the equations for the x and y-component of "r" to obtain the time of flight and the initial velocity:
Using the equation for the y-component of the vector position:
y = y0 + v0y · t + 1/2 · g · t² (y0 and v0y = 0)
-90 m = -1/2 · 9.8 m/s² · t²
-2 · -90 m/ 9.8 m/s² = t²
t = 4.3 s
Using the equation for the x-component of the vector position:
x = x0 + v0x · t
70 m = v0x · 4.3 s
v0x = 16 m/s
The initial velocity is 16 m/s
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where
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which is the distance at which the other wire is located:
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2) direction of the force:
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- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)
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- thumb: force --> due east --> so the force is repulsive
A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.