Ask question

Ask question

All categories

3 years ago

13

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.
[Delayed Feedback]
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

You might be interested in

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

3 years ago

If 1 foot is 30.28 centimeters.how many cm is 130 feet

uranmaximum [27]

Answer:3,936.4

Explanation:

If 1 foot is 30.28 centimeters multiply 30.28 by 130

6 0

3 years ago

Three noise sources produce volume (loudness) levels of 70, 73, and 80 dB when acting separately. When the sources act together,

Mashutka [201]

Sound at 70 dB is 70 dB louder than the human reference level. That's 10⁷ times as much as the reference sound power.

Sound at 73 dB is 73 dB louder than the human reference level. That's 10⁷.³ or 2 x 10⁷ times as much as the reference sound power.

Sound at 80 dB is 80 dB louder than the human reference level. That's 10⁸ or 10 x 10⁷ times as much as the reference sound power.

Now we can adumup:

Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)

Intensity = (13 x 10⁷) times the sound power reference intensity.

Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)

Intensity = 70 + 10 log(13)

Intensity = 70 + 10 (1.114)

Intensity = 70 + 11.14

Intensity = <em>81.14 dB</em>

<em>______________________________________</em>

Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct. For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.

8 0

3 years ago

The rate of change of velocity is called:.

Pavel [41]

Answer: Acceleration

Detailed Explanation:

Acceleration is defined as the rate of change of velocity.

6 0

2 years ago

Which has more thermal energy: lake or a cup of hot chocolate?

lawyer [7]

Though the hot cocoa would have a higher temperature, the lake would have more thermal energy because it has more molecules with a greater total internal energy.

4 0

3 years ago