Physics, 21.07.2019 11:30 hannahkharel2

Agirl throws a rock horizontally, with a velocity of 10 m/s, from a bridge. it falls 20 m to the water below. how far does the rock travel horizontally before striking the water, assuming negligible air resistance?

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Before you begin, you need to first figure out what your problem gives you and what you need to look for.
Given:
vix=10m/s
dy= 20m

What you are looking for is horizontal distance or dx.

The formula for vertical distance is:
$dx=vi_{x}t$

Where:
dx= horizontal distance
vix= initial horizontal velocity
t=time

But when you look at your problem, you do not have time. Now for time, you need to use the formula:

$t= \sqrt{ \frac{2dy}{g} }$

g in this equation is a constant. It is the acceleration due to gravity which has the value of 9.8m/s^2.

Now we first need to get time:
$t= \sqrt{ \frac{2dy}{g} }$
$t= \sqrt{ \frac{2(20m}{9.8m/s^{2}} }$
$t= \sqrt{ 4.08163265}$
t=2.02s

Now that you have time, all you need to do is insert that into our equation for horizontal distance.

$dx=vi_{x}t$
dx=(10m/s)(2.02s)

The horizontal distance of the rock will be 20.20m.

Answer from: Quest

the ima (higher than 1m) will be greater than the ima with the height of 1m.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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Agirl throws a rock horizontally, with a velocity of 10 m/s, from a bridge. it falls 20 m to the wat...

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