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pychu [463]
3 years ago
12

If you can throw a stone straight up to height h, what’s the maximum horizontal distance you could throw it over level ground?

Physics
2 answers:
Mariana [72]3 years ago
8 0
To answer this question, first we take note that the maximum height that can be reached by an object thrown straight up at a certain speed is calculated through the equation,
                            Hmax = v²sin²θ/2g
where v is the velocity, θ is the angle (in this case, 90°) and g is the gravitational constant. Since all are known except for v, we can then solve for v whichi s the initial velocity of the projectile. 

Once we have the value of v, we multiply this by the total time traveled by the projectile to solve for the value of the range (that is the total horizontal distance). 
vovikov84 [41]3 years ago
4 0

Answer:

R = \frac{u^2}{g}

Explanation:

Maximum distance covered by the stone can be calculated as follows:

R = \frac{u^2 sin2\theta}{g}

where, u is the initial velocity, θ is the launch angle and g is the acceleration due to gravity.

For maximum horizontal distance, θ should be 45°. Then maximum distance covered would be:

R = \frac{u^2}{g}

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Mandarinka [93]

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In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

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3 years ago
Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it
MakcuM [25]

Answer:

Explanation:

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Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,
Nutka1998 [239]
Good morning.

We have:

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Where j is the unitary vector in the direction of the y-axis.

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Therefore, the magnitude of b is 47 units.
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