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mamaluj [8]
2 years ago
15

Guys please helpp!!!!1

Physics
1 answer:
Setler79 [48]2 years ago
3 0

Answer:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

Explanation:

Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy (E) is the sum of gravitational potential energy (U) and translational kinetic energy (K), all measured in joules. In addition, gravitational potential energy is directly proportional to height (h) and translational kinetic energy is directly proportional to the square of velocity.

Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:

Position A/Position E

K = E, U = 0

Position B/Position D

K = (1-x)\cdot E, U = x\cdot E, for 0 < x < 1

Position C

K = 0, U = E

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The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




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