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8. A golfer hits a ball and give it an initial velocity of 25 m/s, at an angle of 35° above the horizontal.

a) How long downs the ball stay in the air?

b) How far horizontally (the range) does the ball travel before hitting the ground?​

Respuesta :

Answer:

b) How far horizontally (the range) does the ball travel before hitting the ground?

MDKPfromIDN

So, here is your answer :

  • [tex] \sf{t_{x \: max}} [/tex] = time to reach the farthest point ≈ 2.93 s
  • [tex] \sf{x_{max}} [/tex] = the farthest point of parabolic motion ≈  59.94 m

Introduction :

Hi ! I will help you with the dynamics of parabolic motion. Remember that a parabolic motion will be impossible to happen if it doesn't have a certain elevation angle (due to the vertical and horizontal axis). Almost any equation that related to parabolic motion is affected by the initial velocity, initial angle of elevation (especially to the horizontal axis), and the acceleration due to gravity at a place. In this introduction, I will add formulas to calculate the time it reaches the furthest point and the farthest point traveled by an object.

Formula Used :

Time to stay in the air (time to reach the farthest point)

[tex] \boxed{\sf{\bold{t_{x \: max} = \frac{2 \times v_0 \times \sin(\theta)}{g}}}} [/tex]

With the following condition:

  • [tex] \sf{t_{x \: max}} [/tex] = time to reach the farthest point (s)
  • [tex] \sf{v_0} [/tex] = initial speed (s)
  • [tex] \sf{\theta} [/tex] = angle elevation ([tex] \:^o [/tex] atau rad)
  • [tex] \sf{g} [/tex] = acceleration of the gravity (m/s²)

The farthest point of parabolic motion

[tex] \boxed{\sf{\bold{x_{max} = \frac{(v_o)^2 \times \sin(2\theta)}{g}}}} [/tex]

With the following condition:

  • [tex] \sf{x_{max}} [/tex] = the farthest point of parabolic motion (m)
  • [tex] \sf{v_0} [/tex] = initial speed (s)
  • [tex] \sf{\theta} [/tex] = angle of elevation ([tex] \:^o [/tex] atau rad)
  • [tex] \sf{g} [/tex] = acceleration of the gravity (m/s²)

Problem Solving

We know that:

  • [tex] \sf{v_0} [/tex] = initial speed = 25 m/s
  • [tex] \sf{\theta} [/tex] = angle of elevation = 35°
  • [tex] \sf{g} [/tex] = acceleration of the gravity = 9.8 m/s²

What was asked :

  1. [tex] \sf{t_{x \: max}} [/tex] = time to reach the farthest point = ... s
  2. [tex] \sf{x_{max}} [/tex] = the farthest point of parabolic motion = ... m

Step by step :

[1] [tex] \sf{\bold{t_{x \: max}}} [/tex] = ... s

[tex] \sf{t_{x \: max} = \frac{2 \times v_0 \times \sin(\theta)}{g}} [/tex]

[tex] \sf{t_{x \: max} = \frac{2 \times 25 \times \sin(35^o)}{9.8}} [/tex]

[tex] \sf{t_{x \: max} = \frac{50 \times 0.574}{9.8}} [/tex]

[tex] \sf{t_{x \: max} = \frac{28.7}{9.8}} [/tex]

[tex] \boxed{\sf{t_{x \: max} \approx 2.93 \: s}} [/tex]

______

[2] [tex] \sf{\bold{t_{x \: max}}} [/tex] = ... s

[tex] \sf{x_{max} = \frac{(v_o)^2 \times \sin(2\theta)}{g}} [/tex]

[tex] \sf{x_{max} = \frac{(25)^2 \times \sin(2 \cdot 35^o)}{9.8}} [/tex]

[tex] \sf{x_{max} = \frac{625 \times \sin(70^o)}{9.8}} [/tex]

[tex] \sf{x_{max} = 63.77 \times 0.94} [/tex]

[tex] \boxed{\sf{x_{max} \approx 59.94 \: m}} [/tex]

Conclusion :

  1. [tex] \sf{t_{x \: max}} [/tex] = time to reach the farthest point ≈ 2.93 s
  2. [tex] \sf{x_{max}} [/tex] = the farthest point of parabolic motion ≈  59.94 m