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Elden
3 months ago
11

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.
Physics
1 answer:
Maru [3.3K]3 months ago
0 0

Answer:

A)    t = 0.40816 s, y = 0.916 m

Explanation:

A) For solving this problem, we utilize kinematic equations.

           v = v₀ - g t

For the highest point, the velocity is zero (v = 0).

           t = (v₀ - v) / g

           t = (4 - 0) / 9.8

           t = 0.40816 s

To calculate the height, we apply the equation

          v² = v₀² - 2 g y

           y = v₀² / (2g)

           y = 4² / (2 * 9.8)

           y = 0.916 m

To ascertain the number of photos, we can apply a direct proportionality method: if you take 30 photos in a second for 0.40816 s, how many photos are taken?

          # _photos1 = 0.40816 (30/1)

          # _photos1 = 12

Yes, I capture at 120 fps.

          #_fotod = 0.40816 (120/1)

          #photos = 5.87 * 10³

 

B) The ball is dropped from height h; we want to find the time it takes to reach the ground.

           v² = v₀² + 2 g y

where the initial velocity is zero, and the velocity the expert releases is equal to the outgoing speed v = 4 m/s.

            v² = 2gy

             v = √(2 * 9.8 * 0.916)

             v = √(2.1397 * 101)

             v = 4.6257 m/s

C) We need to calculate the two instances when the ball reaches a height of y = ½ h. The final formulation and numerical outcomes will be provided.

Now applying the formula

             y = L / 2

             y = 0.916 / 2

             y = 0.458 m

At this point, we can apply the equation

             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 need to solve the quadratic equation.

           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 period is when the ball ascends, while the longest is during its descent.

D) The expected plot of vs versus time is anticipated to form a closed loop, while the position against time should resemble a parabola.

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The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. how slowly can the ball go through its top
Keith_Richards [3271]
<span>A centripetal force maintains an object's circular motion. When the ball is at the highest point, we can assume that the ball's speed v is such that the weight of the ball matches the required centripetal force to keep it moving in a circle. Hence, the string will not become slack. centripetal force = weight of the ball m v^2 / r = m g v^2 / r = g v^2 = g r v = sqrt { g r } v = sqrt { (9.80~m/s^2) (0.7 m) } v = 2.62 m/s Thus, the minimum speed for the ball at the top position is 2.62 m/s.</span>
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3 months ago
You throw a baseball at an angle of 30.0∘∘ above the horizontal. It reaches the highest point of its trajectory 1.05 ss later. A
kicyunya [3294]

Answer:

The initial speed at which the baseball is thrown is 20.58 m/s

Explanation:

To determine the time taken to reach the peak height in projectile motion, the formula is:

t = (u sin θ)/g

Where u = initial speed of the baseball =?

θ = angle of launch above the horizontal

g = gravitational acceleration = 9.8 m/s²

1.05 = (u sin 30)/9.8

u can be calculated as (1.05 × 9.8)/0.5

u = 20.58 m/s

7 0
2 months ago
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