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3 years ago

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A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w

hat time th did that take? Show the final formulas for the quantities, along with the numerical answers. At 30 fps how many frames would be recorded in a video of the throw? At 120 fps?
B. From the height h in slide 2 the ball then drops back to y = 0. Find the time t0 that takes, and the velocity there (magnitude and direction). Show the final formulas for the quantities, along with the numerical answers.
C. Find the two times the ball is at the height y = ½ h. Show the final formula and the numerical answers.
D. Roughly sketch on a piece of paper what you expect the plots of y versus t and vy versus t will look like, take a picture, and paste them here.

1 answer:

sveta [45]3 years ago

0 0

Answer:

A) t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

v = v₀ - g t

the highest point zero velocities (v = 0)

t = (v₀-v) / g

t = (4 - 0) / 9.8

t = 0.40816 s

to calculate the height let's use

v² = v₀² - 2 g y

y = vo2 / 2g

y = 4 2 / (2 9.8)

y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

# _photos1 = 0.40916 (30/1)

# _photos1 = 12

yes i take 120 fps

#_fotod = 0.40916 (120/1)

#photos = 5.87 10³

B) The ball is released from a latura h how long it takes to reach the floor

v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

v² = 2gy

v = √ (2 9.8 0.916)

v = √ (2.1397 101)

v = 4.6257 m / s

c) we ask us for the time for latura

y = L / 2

y = 0.916 / 2

y = 0.458 m

now we can use the formula

y = v₀ t - ½ g t²

0.458 = 4.00 t - ½ 9.8 t²

4.9 t² - 4t + 0.458 = 0

t² -0.8163 t +0.09346 = 0

we solve second degree execution

t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

t = [0.8163 ± 0.540] / 2

t₁ = 0.678 m

t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

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From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

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$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

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b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

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$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

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