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shepuryov [24]
3 years ago
14

9. The maximum horizontal range of a projectile is given by the formula R= u2/g where u is the initial velocity and g is the acc

eleration due to gravity. Find the velocity with which a ball can be thrown to have a maximum range of 20 meters when the acceleration due to gravity is equal to 9.8 m/s.
(SHOW WORK)
Mathematics
2 answers:
muminat3 years ago
5 0

Answer:

<em>The velocity with which a ball must be thrown to have a maximum range of 20 m </em><u>is 14 m/s.</u>

Note that this problem means to find the magnitude of the velocity and not the direction (it is implicit in the formula that the angle of the launch is 45°).

Explanation:

You just must use the given equation for the maximum horizontal range of a projectile and solve for u which is the unknwon:

  • Given equation: R = u² /g
  • g = 9.8 m/s²
  • R = 20 m
  • u =?

Solve for u:

  • u² = R × g = (20 m) × (9.8 m/s²)  =  196 m²/s²

Take square root from both sides:

  • u = 14 m/s ← answer
Helen [10]3 years ago
4 0
<h2>Hello!</h2>

The answer is:

The velocity with which the ball can be thrown to have a maximum range of 20 meters is equal to 14 m/s.

u=14\frac{m}{s}

<h2>Why?</h2>

To solve the problem and find the velocity, we need to isolate it from the equation used to calculate the maximum horizontal range.

We have the equation:

R=\frac{u^{2} }{g}

Where,

R is the maximum horizontal range.

u is the initial velocity.

g is the gravity acceleration.

Also, from the statement we know that:

R=20m\\g=9.8\frac{m}{s^{2} }

So, using the given information, and isolating, we have:

R=\frac{u^{2} }{g}

R*g=u^{2}

u^{2}=R*g=20m*9.8\frac{m}{s^{2} }=196\frac{m^{2} }{s^{2} }\\\\u=\sqrt{196\frac{m^{2} }{s^{2}}}=14\frac{m}{s}

Hence, we have that the velocity with which the ball can be thrown to have a maximum range of 20 meters is equal to 14 m/s.

u=14\frac{m}{s}

Have a nice day!

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<h3>What is a percentage?</h3>

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