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Degger [83]
3 years ago
6

A skier (m=59.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 3

.00 m and D = 13.9 m, find H. Submit Answer Tries 0/12 Find her total kinetic energy as she reaches the ground.
Physics
1 answer:
marissa [1.9K]3 years ago
4 0

Answer:

35.20 m

Explanation:

By the law of conservation of energy we have,

mg(H-h)=\frac{1}{2}mv^2

g(H-h)=\frac{1}{2}v^2

\Rightarrow H=\frac{v^2}{2g}+h

where m= mass of the skier, h= 3.00 m

D= horizontal distance=13.9 m

H= maximum height attained

Also, the horizontal distance covered by the skier is

D=vt

=v\frac{2g}{h}

\Rightarrow v^2=\frac{gD^2}{2h}

thus, height H in terms of D  is given by

H=\frac{D^2}{2h}+h

H=\frac{13.9^2}{2\times3}+3

H=35.20 m

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vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses
dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}

(This is correct because the horizontal motion has acceleration zero). Then:

v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}

Then, plugging in the given values, we obtain:

k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m

Finally, the effective spring constant of the firing mechanism is 1808N/m.

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Answer:

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