With what initial velocity must an object be thrown upward from a height of 2 meters to reach a maximum hieght of 200 meters

1 answer:

6 0

The equations of motion to use for a constant acceleration a = g = -9.81:
$v = gt + v_0$
$y = \frac{1}{2} g t^{2} + v_0t + y_0$

at the highest point the velocity must be zero:
$0 = gt + v_0 \\ v_0 = -gt$
combine both equations:
$y = \frac{1}{2} \frac{v_0^{2} }{g} - \frac{v_0^{2} }{g} + y_0= -\frac{1}{2} \frac{v_0^{2} }{g} + y_0$
Solve for v₀:
$v_0 = \sqrt{2g(y_0 - y)}$

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I am struggling on this physics question. Brainly is my last hope. Could somebody please provide an answer to this question, wit

Aleks [24]

1) 29.4 N

The force of gravity between two objects is given by:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

M and m are the masses of the two objects

r is the separation between the centres of mass of the two objects

In this problem, we have

$M=5.97\cdot 10^{24} kg$ (mass of the Earth)

$m=3.0 kg$ (mass of the box)

$r=R=6.37\cdot 10^6 m$ (Earth's radius, which is also the distance between the centres of mass of the two objects, since the box is located at Earth's surface)

Substituting into the equation, we find F:

$F=\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24})(3.0)}{(6.37\cdot 10^6)^2}=29.4 N$

2) $g=9.8 m/s^2$

Let's now calculate the ratio F/m. We have:

F = 29.4 N

m = 3.0 kg

Subsituting, we find

$\frac{F}{m}=\frac{29.4}{3.0}=9.8 N/kg = 9.8 m/s^2$

This is called acceleration of gravity, and it is the acceleration at which every object falls near the Earth's surface. It is indicated with the symbol $g$ .

We can prove that this is the acceleration of the object: in fact, according to Newton's second law,

$F=ma$

where a is the acceleration of the object. Re-arranging,

$a=\frac{F}{m}$

which is exactly equal to the quantity we have calculated above.

5 0

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A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be for a

andriy [413]

To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

$sin\theta = 1.22\frac{\lambda}{d}$