contestada

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/6. In terms of R and g, find (a) the time the ball is in motion, (b) the ball’s speed at the peak of its path, (c) the initial vertical component of its velocity, (d) its initial speed, and (e) the angle i . (f) Suppose the ball is thrown at the same initial speed found in part (d) but at the angle appropriate for reaching the maximum height. Find this height. (g) Sup- pose the ball is thrown at the same initial speed but at the angle necessary for maximum range. Find this range.

Respuesta :

Answer:

Part a)

[tex]T = 2\sqrt{\frac{R}{3g}}[/tex]

Part b)

[tex]v_x = \frac{\sqrt{3Rg}}{2}[/tex]

Part c)

[tex]v_y = \sqrt{Rg/3}[/tex]

Part d)

[tex]v = \frac{1}{2}\sqrt{13Rg}[/tex]

Part e)

[tex]\theta_i = 33.7 degree[/tex]

Part f)

[tex]H = \frac{13R}{8}[/tex]

Part g)

[tex]X = \frac{13R}{4}[/tex]

Explanation:

Initial speed of the launch is given as

initial speed = [tex]v_i [/tex]

angle = [tex]\theta_i[/tex] degree

Now the two components of the velocity

[tex]v_x = v_i cos\theta_i[/tex]

similarly we have

[tex]v_y = v_i sin\theta_i[/tex]

Part a)

Now we know that horizontal range is given as

[tex]R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}[/tex]

maximum height is given as

[tex]H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}[/tex]

so we have

[tex]v_i sin\theta = \sqrt{Rg/3}[/tex]

time of flight is given as

[tex]T = \frac{2v_isin\theta_i}{g}[/tex]

[tex]T = \frac{2\sqrt{Rg/3}}{g}[/tex]

[tex]T = 2\sqrt{\frac{R}{3g}}[/tex]

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

[tex]R = v_x T[/tex]

[tex]R = v_x 2\sqrt{\frac{R}{3g}}[/tex]

[tex]v_x = \frac{\sqrt{3Rg}}{2}[/tex]

Part c)

Initial vertical velocity is given as

[tex]v_y = v_i sin\theta_i[/tex]

[tex]v_i sin\theta = \sqrt{Rg/3}[/tex]

Part d)

Initial speed is given as

[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

so we will have

[tex]v = \sqrt{Rg/3 + 3Rg/4}[/tex]

[tex]v = \frac{1}{2}\sqrt{13Rg}[/tex]

Part e)

Angle of projection is given as

[tex]tan\theta_i = \frac{v_y}{v_x}[/tex]

[tex]tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}[/tex]

[tex]\theta_i = 33.7 degree[/tex]

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

[tex]H = \frac{v^2}{2g}[/tex]

[tex]H = \frac{13R}{8}[/tex]

Part g)

For maximum range the angle should be 45 degree

so maximum range is

[tex]X = \frac{v^2}{g}[/tex]

[tex]X = \frac{13R}{4}[/tex]

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