Physics, 26.09.2019 16:10 queengenni

26. a ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) at what speed does the ball hit the ground? (b) for how long does the ball remain in the air? (c)what maximum height is attained by the ball?

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Answer from: liln223

The given problem models a parabolic motion, where the intial speed is the same as the final speed when the ball hits the ground, assuming that there's no wind.

So, by given we already know the horizontal and vertical initial velocities, which are

$v_{o_{y} }=12m/s\\ v_{o_{x} }=16m/s$

Now, to find the initial velocity (which is equal to the velocity when hits the ground), we need to sum both vector magnitues with the following formula

$v_{o}=\sqrt{(v_{o_{x} } )^{2}+(v_{o_{y} } )^{2} } \\v_{o}=\sqrt{(16)^{2} +(12)^{2} }=\sqrt{256+144} =\sqrt{400}=20m/s\\\therefore v_{o}=v_{f}=20m/s$

To find the time thorugh which the ball remains in the air, we need to find the times that took to reach the highest point, and then we duplicate it, because the movement is vertically symmetrical. So, the tima to the highest point is defined as

$t=\frac{v_{o_{y} } }{g}$

Where, $v_{o_{y} }=12m/s$ and $g=9.81 m/s^{2}$ .

(Remember that we need to use the vertical initial velocity, beacuse the times to get to the highest points depends on vertical magnitudes).

Replacing all values, we have

$t=\frac{v_{o_{y} } }{g}=\frac{12m/s}{9.81 m/s^{2} }\\ t \approx 1.22 sec$

So, the time while the ball was in the air is

2t=2(1.22sec)=2.44sec

The last question is about the highest point the ball reached, that can be found with

$h_{max}=\frac{v_{o_{y} }^{2} }{2g}$

Replacing all values, we have

$h_{max}=\frac{v_{o_{y} }^{2} }{2g}=\frac{(12m/s)^{2} }{2(9.81m/s^{2}) }\\h_{max}=\frac{144}{19.62}m \approx 7.3 m$

Therefore,

The speed of the ball when hits the ground is 20m/s.The time thorugh which the ball remained in the air is 1.22 seconds.The maximum height that the ball reached is 7.3 meters.

Answer from: Quest

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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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26. a ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in th...

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