Physics, 15.04.2020 03:35 jby

A) An astronaut holds a rock 100 m above the surface of Planet X. The rock is then thrown upward with a speed of 15 m/s, as shown in the figure. The rock reaches the ground 10 s after it is thrown. The atmosphere of Planet X has a negligible effect on the rock when it is in free fall.

i) Determine the acceleration due to gravity of the rock when it is on Planet X.

ii) How does the speed of the rock when it reaches the ground vd compare to the speed of the rock when it is thrown upward vu? State your reasoning.

b) A student wants to know how the motion of the rock would be different if it was thrown upward at 15 m/s from a height of 100 m above Earth s surface. In a clear, coherent, paragraph-length response that may also contain figures and/or equations, explain how the motion of the rock on Earth will be different from its motion on Planet X in terms of its maximum height above the ground, the speed at which it reaches the ground, the time in which it is in free fall, and its acceleration due to gravity.

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Answer from: bcox32314

(i) Acceleration due to gravity on Planet X is 5m/s².

(ii) The speed of the rock when it reaches the ground is 2.3 times its speed when it is thrown upward.

Explanation:

a) Let's divide the motion in three parts to simplify the problem. The first one is the upward motion that comes to a stop at the highest point. The second is the downward motion from the highest point to the initial point at 100m high. the third is the downward motion from that point to the surface of the planet.

Since the vertical motion is symmetrical, we know that the time of the first and second sections of the movement are the same; and the initial speed of the first one is equal to the final speed of the second one. Then, we can calculate the total time of these two parts with the formula:

$t_{1,2}=2t_1=2\frac{v_0}{g_x}$

Where $t_{1,2}$ is the time of the sections 1 and 2, v_0 is the initial speed of the rock and g_x is the acceleration of gravity in Planet X.

Now, the time of the third section of the motion can be calculated with the equation:

$t_3=\frac{v_f-v_0}{g_x}$

So, the total time is given by:

$t=t_{1,2}+t_3=\frac{2v_0+v_f-v_0}{g_x}=\frac{v_0+v_f}{g_x}$

We can solve the final velocity from the equation:

$v_f^{2}=v_0^{2}+2g_xy\\\\\implies v_f=\sqrt{v_0^{2}+2g_xy}$

And substitute in the previous expression:

$t=\frac{v_0+\sqrt{v_0^{2}+2g_xy}}{g_x}$

Now, we solve for g_x :

$g_xt-v_0=\sqrt{v_0^{2}+2g_xy} \\\\g_x^{2}t^{2}-2g_xv_0t+v_0^{2}=v_0^{2}+2g_xy\\\\g_x^{2}t^{2}-2g_xv_0t-2g_xy=0$

At this point, since $g_x\neq 0$ , we can divide and simplify:

$g_xt^{2}-2v_0t-2y=0\\\\\implies g_x=2\frac{v_0t+y}{t^{2}}$

Finally, we plug in the known values and compute:

$g_x=2\frac{(15m/s)(10s)+100m}{100s^{2}}\\\\g_x=5m/s^{2}$

This means that the acceleration due to gravity on Planet X is 5m/s^{2} (i).

To calculate the speed of the rock when it hits the ground, we use the equation above:

$t=\frac{v_0+v_f}{g_x} \\\\v_f=g_xt-v_0\\\\v_f=(5m/s^{2})(10s)-15m/s\\\\v_f=35m/s$

Now, we find the ratio between the initial and final velocity:

$\frac{v_f}{v_0}=\frac{35m/s}{15m/s}=2.3$

It means that the speed of the rock when it reaches the ground is 2.3 times the speed of the rock when it is thrown upward (ii).

b) Since the acceleration due to gravity on Planet X is different than on Earth, the whole motion is modified:

The maximum height above the ground is lower; it can be calculated from:

$y_{max}_x=y_0+\frac{v_0^{2}}{2g_x}=100m+\frac{(15m/s)^{2}}{2(5m/s^{2})}=122.5m\\\\y_{max}_e=y_0+\frac{v_0^{2}}{2g_e} =100m+\frac{(15m/s)^{2}}{2(9.8m/s^{2})}=111.5m$

The speed at which it reaches the ground is higher, since it depends on the acceleration due to gravity:

$v_f_e=\sqrt{v_0^{2}+2g_ey} =\sqrt{(15m/s)^{2}+2(9.8m/s^{2})(100m)}=46.7m/sv_f_x$

The time in which it is in free fall is lower, since the rock is attracted with a greater force to the ground and it falls faster:

$t_e=\frac{v_0+v_f}{g_e}=\frac{15m/s+46.7m/s}{9.8m/s^{2}}=6.3s$

The acceleration due to gravity is nearly twice the value it has in Planet X:

$\frac{g_e}{g_x}=\frac{9.8m/s^{2}}{5m/s^{2}}=1.96\implies g_e=1.96g_x$

Answer from: Quest

explanation:

Answer from: Quest

volts times amps 6 x 0.5 3w

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