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A golfer hits a ball and gives it an initial velocity of 40 m/s, at an angle of 30 degrees above the horizontal.

A) how long does the ball stay in the air? (Answer is 4.08 sec)

B) how far horizontally (the range) does the ball travel before hitting the ground? (Answer 141 m)

I need help with getting the work. If y’all could explain, that would be so helpful and thank you.

A golfer hits a ball and gives it an initial velocity of 40 ms at an angle of 30 degrees above the horizontal A how long does the ball stay in the air Answer is class=

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asuwafohjames

(a) The ball will stay in the air for 4.08 s.

(b) The ball travels 141 m horizontally before hitting the ground.

This is a projectile motion problem, to solve the problems above, we need to use the formula of (a) Time of flight and (b) Range.

(a)

⇒ Formula

  •   T = 2Usin∅/g.................... Equation 1

⇒Where:

  • T = Time of flight
  • U = initial velocity
  • ∅ = Angle above the horizontal
  • g = acceleration due to gravity

From the question,

⇒Given:

  • U = 40 m/s
  • ∅ = 30°

⇒Assuming:

  • g = 9.8 m/s²

⇒Substitute these values into equation 1

  • t = 40sin30/9.8
  • t = 2(40)(0.5)/9.8
  • t = 40/9.8
  • t = 4.08 s

Also,

(b) For Range

⇒ Formula

  • R = U²sin2∅/g............... Equation 2

⇒Substitute these values above into equation 2

  • R = 40²sin(2×30)/9.8
  • R = 1600×0.866/9.8
  • R = 141.38 m
  • R ≈ 141 m

Hence, (a) The ball will stay in the air for 4.08 s (b) The ball travels 141 m horizontally before hitting the ground

Learn more about projectile motion here: https://brainly.com/question/11049671

A. The ball stay in the air for 4.08 seconds.

B. The ball travel nearly 141 meter before hitting the ground.

Given that,  Initial velocity (v) of ball is 40 m/s, at an angle of 30 degrees above the horizontal.

[tex]v=40m/s, \theta=30,g=9.8m/s^{2}[/tex]

Here we use formula,

  [tex]t=\frac{2v*sin\theta}{g}\\\\range(R)=\frac{v^{2}*sin2\theta }{g}[/tex]

Substituting values in above equations.

[tex]t=\frac{2*40*sin30}{9.8}=\frac{2*40*0.5}{9.8}=4.08s\\\\R=\frac{(40)^{2}*sin60 }{9.8} =\frac{1600*0.866}{9.8} =141m[/tex]

Hence, The ball stay in the air for 4.08 seconds and The ball travel nearly 141 meter before hitting the ground.

Learn more:

https://brainly.com/question/3159582

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