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Masja [62]
3 years ago
11

A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio

n of the kick starts running towards the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?
Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Answer:

U = √Rg/sin2θ

Explanation:

Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.

Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.

According to the range formula,

R = U²sin2θ/g

Cross multiplying we have;

Rg = U²sin2θ

Dividing both sides by sin2θ, we have;

U² = Rg/sin2θ

Taking the square root of both sides we have;

√U² = √Rg/sin2θ

U = √Rg/sin2θ

Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ

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Answer:

1)  \rm q_2 is<u> positive.</u>

<u></u>

2) \rm q_2=4.56\times 10^{-10}\ C.

Explanation:

<h2><u>Part 1:</u></h2>

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., \rm q_2 is <u>positive.</u>

<u></u>

<h2><u>Part 2:</u></h2>

<u></u>

<u>Given:</u>

  • Mass of the balloon, m = 0.00275 kg.
  • Charge on the balloon, \rm q_1 = -3.50\times 10^{-8}\ C.
  • Distance between the rod and the balloon, d = 0.0640 m.
  • Acceleration due to gravity, \rm g = 9.81\ m/s^2.

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, \rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.

The magnitude of the electrostatic force on the balloon due to the rod is given by

\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.

\rm \dfrac{1}{4\pi \epsilon_o} is the Coulomb's constant.

For the elecric force and the weight to be balanced,

\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.

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