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Ivahew [28]
4 years ago
5

You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa

s the platform, and how far from the edge of the platform did you hit the water? Assume your initial velocity is horizontal. Ignore air resistance.
Physics
1 answer:
rusak2 [61]4 years ago
7 0

The distance you free-fall from rest is  D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change.  It's constant.  Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

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Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the
Airida [17]

Complete Question:

Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

V =\frac{Kq}{r}

where;

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r is the distance of the charge

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The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V

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