NEED HELP ASAP PLS (with work)
A punt is kicked at an angle of 60 with a speed of 40 m/s, and lands at the same
height from which it was kicked.

How far away will the ball land?

What is the maximum height reached by the ball?

What is the hang time? Total

Answers

Answer 1

Answer:

1) The ball will land approximately 141.4 m away from where it was kicked

2) The maximum height reached is approximately 61.22 m

3) The total hang time is approximately 7.07 seconds

Explanation:

First we list out the known parameters

The direction in which the punt is kicked, θ = 60°

The speed with which the punk is kicked, v₀ = 40 m/s

1) The distance away (the range, R) the ball will land is given by the equation;

[tex]R = \dfrac{v_0^2 \times sin(2\cdot \theta)}{g}[/tex]

Where;

g =  The acceleration due to gravity = 9.8 m/s²

Substituting the values, gives;

[tex]R = \dfrac{40^2 \times sin(2\times60^{\circ })}{9.8} \approx 141.4 \ m[/tex]

The distance away the ball will land ≈ 141.4 m

2) The maximum height is given m=by the following equation for vertical motion;

(v₀ × sin(θ))² = 2 × g × [tex]h_{max}[/tex]

Obtained from the equation of motion, v² = u² - 2×g×h where at maximum height;

v = 0 at maximum height, the ball stops upward motion

h = Height of motion = [tex]h_{max}[/tex] at maximum height

v, and u are the vertical component of the velocity, v₀

Substituting the values in the equation for the maximum height, gives;

(40 × sin(60))² = 2 × 9.8 × [tex]h_{max}[/tex]

(40 × (√3)/2)² = 1200 = 19.6 × [tex]h_{max}[/tex]

1200 = 19.6 × [tex]h_{max}[/tex]

[tex]h_{max}[/tex] = 1200/19.6 ≈ 61.22 m

The maximum height reached ≈ 61.22 m

3) The time to maximum height is given by the following equation of motion;

[tex]v_y[/tex] = v₀ × sin(θ) - g × t

[tex]v_y[/tex] = 0 at maximum height

We therefore have;

v₀ × sin(θ) = g × t

t =  v₀ × sin(θ)/g = 40 × sin(60°)/9.8 = (40 ×(√3)/2)/9.8 = 3.535 seconds

Given that the motion of the ball is symmetrical about the maximum height, we have;

The time from ground level to maximum height = The time from the maximum height back to the ground level

The total time hang time = The time from ground level to maximum height + The time from the maximum height back to the ground level

The total time hang time ≈ 3.535 + 3.535 ≈ 7.07 seconds.

The total time hang time ≈ 7.07 seconds.


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A water-skier is pulled behind a boat by a rope that is at an angle of 13° and has a tension of 490 N. The water-skier has a mass of 49 kg.

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Answers

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Explanation:

I'm sorry I just did this today in class but i think normal force has to be equal to mg or the mass * gravity of the person and that would be 480.2 newtons. I multiplied 9.8 (gravity) by the mass of 49g

I think for the x and y components you can make a triangle with the angled string and then use sohcahtoa to solve for the tension. I used cos(13) = x/490 to solve for x component and then I used the pythagoran theroem to get the remaining side which would be : a^2 + 477.4^2 = 490^2

hope this can help and that it is correct. good luck

The magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is 480.2N

The magnitude of the x- and y- components of the tension is expressed as:

x = Tcos 13°y = T sin  13°

Given the following

Tension T = 490N

x = 490 cos13° = 477.44N

y = 490sin13° =  110.23N

Hence the magnitudes of the x- and y- components of the tension is 477.44N and 110.23N

The normal force acting on the skier is equal to the weight.

N = W = mg

N = W = 49 * 9.8

N = W = 480.2N

Hence the normal force acting on the skier is 480.2N

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