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prisoha [69]
3 years ago
15

During a tennis match, a player serves the ball at a speed s, with the center of the ball leaving the racquet at angle θ below t

he horizontal at a height y0. The net is a distance d away and has a height h. When the ball is directly over the net, what is the distance between them?
Physics
1 answer:
Pachacha [2.7K]3 years ago
5 0

Answer: Maximum distance

= {s²/g} * sine(2*theta)unit

Explanation: This is a projectile motion problem. The horizontal distance between the tennis player and where the tennis reaches over the net is given by the horizontal Range.

Range = {s² * sine2*theta}/g

(s)is the initial speed of projection

Theta is the angle of projection

g is acceleration due to gravity 10m/s².

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The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
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Answer:

 y = 17 m

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          x = v₀ₓ t

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          45 = v₀ cos θ t

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         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

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Let's use the last two

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we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

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        t = √ (10 / 4.9)

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Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

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         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

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the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

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