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3 years ago

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Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

ed of 24 m/s at an angle 55 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.
1)What is the horizontal component of the ball’s velocity when it leaves Julie's hand?m/s
2)What is the vertical component of the ball’s velocity when it leaves Julie's hand?
)What is the maximum height the ball goes above the ground?m
)What is the distance between the two girls?
5)After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 15 m/s when it reaches a maximum height of 24 m above the ground.
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6)How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)m

2 answers:

Lisa [10]3 years ago

7 0

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒ 0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0 and x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If v₀x = 15 m/s = vx and ymax = 24 m

y = ? when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒ y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒ y = 18.4 m

wel3 years ago

5 0

Answer:

1) The horizontal component is 13.77 m/s

2) The vertical component is 19.66 m/s

3) The maximum height is 21.2 m

4) The distance between the two girls is 82.07 m

5) No question here

6) The high above the ground is -30.33 m (hit the ground already)

Explanation:

please look at the solution in the attached Word file

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