Respuesta :
Answer:
(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:
[tex]t_{1,2}=2t_1=2\frac{v_0}{g_x}[/tex]
Where [tex]t_{1,2}[/tex] is the time of the sections 1 and 2, [tex]v_0[/tex] is the initial speed of the rock and [tex]g_x[/tex] is the acceleration of gravity in Planet X.
Now, the time of the third section of the motion can be calculated with the equation:
[tex]t_3=\frac{v_f-v_0}{g_x}[/tex]
So, the total time is given by:
[tex]t=t_{1,2}+t_3=\frac{2v_0+v_f-v_0}{g_x}=\frac{v_0+v_f}{g_x}[/tex]
We can solve the final velocity from the equation:
[tex]v_f^{2}=v_0^{2}+2g_xy\\\\\implies v_f=\sqrt{v_0^{2}+2g_xy}[/tex]
And substitute in the previous expression:
[tex]t=\frac{v_0+\sqrt{v_0^{2}+2g_xy}}{g_x}[/tex]
Now, we solve for [tex]g_x[/tex]:
[tex]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[/tex]
At this point, since [tex]g_x\neq 0[/tex], we can divide and simplify:
[tex]g_xt^{2}-2v_0t-2y=0\\\\\implies g_x=2\frac{v_0t+y}{t^{2}}[/tex]
Finally, we plug in the known values and compute:
[tex]g_x=2\frac{(15m/s)(10s)+100m}{100s^{2}}\\\\g_x=5m/s^{2}[/tex]
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:
[tex]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[/tex]
Now, we find the ratio between the initial and final velocity:
[tex]\frac{v_f}{v_0}=\frac{35m/s}{15m/s}=2.3[/tex]
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:
[tex]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<y_{max}_x[/tex]
- The speed at which it reaches the ground is higher, since it depends on the acceleration due to gravity:
[tex]v_f_e=\sqrt{v_0^{2}+2g_ey} =\sqrt{(15m/s)^{2}+2(9.8m/s^{2})(100m)}=46.7m/s>v_f_x[/tex]
- 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:
[tex]t_e=\frac{v_0+v_f}{g_e}=\frac{15m/s+46.7m/s}{9.8m/s^{2}}=6.3s<t_x[/tex]
- The acceleration due to gravity is nearly twice the value it has in Planet X:
[tex]\frac{g_e}{g_x}=\frac{9.8m/s^{2}}{5m/s^{2}}=1.96\implies g_e=1.96g_x[/tex]
The motion of an object in free fall is a motion that is under the effect of only gravitational force
i) The acceleration due to gravity on Planet X is 5 m/s²
ii) The speed of the rock just before it reaches the ground on Planet X is 35 m/s
b) The rock reaches a lower maximum height, but it travels faster and spends less time in free fall on Earth than on Planet X
The reasons the above values are correct are as follows:
The known information are;
Height at which the astronaut throws the rock, h = 100 m
Speed with which the rock is thrown, u = 15 m/s
The time after which the rock reaches the ground, Δt = 10 s
The effect of the atmosphere on Planet X = Negligible
i) Required:
The acceleration due to gravity on Planet X
Solution:
The equation of motion is presented as follows;
s = h + u·t - (1/2)·a·t²
Where;
When the rock reaches the ground, s = 0, t = Δt, therefore;
0 = 100 + 15 × 10 - (1/2)×a×(10)²
a = (100 + 15 × 10)/((1/2)×(10)²) = 5
The acceleration due to gravity on Planet X, a = 5 m/s²
ii) Required:
The comparison between the speed of the rock when it reaches the ground and the speed of the roc when it was thrown upwards
Solution:
The speed of the rock when it reaches the ground, is given as follows;
The maximum height reached, [tex]h_{max} = h + \dfrac{u^2}{2 \cdot a}[/tex]
Therefore;
[tex]h_{max} = 100 + \dfrac{15^2}{2 \times 5} = 122.5[/tex]
The speed at ground, [tex]v_{ground}[/tex] = √(2 × 5 × 122.5) = 35
The speed just before it reaches the ground, [tex]v_{ground}[/tex] = 35 m/s, is more than twice the speed with which the rock is thrown upwards
b) Required:
A paragraph that describes the difference between the motion of the rock on Earth and on Planet X
Solution:
The time the rock takes to reach the ground on Planet X, [tex]t_{ground}[/tex], is given as follows;
v = u - a·t
t = (u - v)/a
Where, v = 0 at the maximum height, we have;
Time to maximum height, [tex]t_{max}[/tex] = 15/5 = 3 seconds
On the way downwards, we have;
v = u + a·t
Where, u = 0, at the maximum height, and [tex]v_{ground}[/tex] = 35 m/s, therefore;
[tex]t_{ground}[/tex] = [tex]v_{ground}[/tex]/a
[tex]t_{ground}[/tex] = 35/5 = 7
The total time of free fall (both on the way upwards and downwards) on Planet X, [tex]t_{tot}[/tex] = 10 m/s
On hearth, we have;
The maximum height reached on Earth, [tex]h_{max} = 100 + \dfrac{15^2}{2 \times 9.8} \approx 111.5[/tex]
On Earth, [tex]v_{ground}[/tex] = √(2 × 9.8 m/s² × 111.5 m) ≈ 46.7 m/s
The time of flight on Earth, [tex]t_{tot}[/tex] = 15/9.8 + 46.7/9.8 ≈ 6.3
The time of flight on Earth, [tex]t_{tot}[/tex] ≈ 6.3 s
Therefore, we have;
The motion of the rock on Earth is such that the maximum height reached on Earth, 111.5 meters, is lesser than the maximum height reached on Planet X, which is 122.5 meters. The rock is travelling faster (approximately 46.7 m/s) when it reaches the ground on Earth compared to the speed of the rock when it reaches the ground on Planet X, where it is travelling at 35 m/s, although the rock spends less time in free fall on Earth (Total time of approximately 6.3 seconds) than the time the rock spends in free fall on Planet X which was 10 seconds
Learn more about the equations of motion in free fall here:
https://brainly.com/question/17610229
