Respuesta :
Answer:
The time taken by the object to reach the ground, t = 25.04 / √gₓ
The final velocity of the object, v = 23.39 √gₓ
Explanation:
Given data,
The height of the Dump Tower, h = 96 x 3.05
= 292.8 m
The mass of the planet housing Dump Tower, M = 2.7 mₓ
The radius of the planet housing Dump Tower, R = 1.7 rₐ
The acceleration due to the gravity of the planet is,
g = GM/R²
= 2.7 Gmₓ / (1.7 rₐ)²
= 0.934 Gmₓ/rₐ²
g = 0.934 gₓ
Using the II equations of motion,
S = ut + ½ gt²
= 0 + ½ (0.934 gₓ) t²
t = √(2S/ 0.934 gₓ)
= √(2 x 292.8/ 0.934 gₓ)
t = 25.04 / √gₓ
Hence, the time taken by the object to reach the ground, t = 25.04 / √gₓ
Using the I equations of motion
v = u + gt
= 0 + 0.934 gₓ (25.04 / √gₓ)
v = 23.39 √gₓ
Hence, the final velocity of the object, v = 23.39 √gₓ
The time is taken by the object to reach the ground and final velocity are 25.04/√gₓ and 23.39√gₓ respectively.
What is kinematics?
It is the study of the motion of a body without considering mass. And the reason behind them is neglected.
Dump Tower is 96 stories tall. A small, 1.2-kg object is dropped over the side of the roof of the tower and accelerates toward the ground.
The mass of the planet housing Dump Tower is 2.7 times the mass of Ganymede and the radius of the body is 1.7 times the radius of Makemake.
You will track the object for its entire fall. Each story of this tower is 3.05 meters tall.
h = 96 × 3.05 = 292.8 m
M = 2.7 mₓ
R = 1.7 rₐ
The acceleration due to gravity will be
[tex]\rm a = \dfrac{GM}{r^2}\\\\\\a = \dfrac{2.7Gm_x}{(1.7*r_a)^2}\\\\\\a = 0.934 *\dfrac{Gm_x}{r_a^2}\\\\\\a = 0.934 * g_x[/tex]
By the second equation of motion, we have
[tex]\rm S = ut + \dfrac{1}{2} at^2\\\\\\h = 0 + \dfrac{1}{2}*0.934 * g_x*t^2\\\\\\t = \sqrt{\dfrac{2h}{0.934*g_x}}\\\\\\t = \dfrac{25.04 }{\sqrt{g_x}}[/tex]
Then the final velocity will be
[tex]\rm v = u + at\\\\v = 0 + 0.934*g_x * \dfrac{25.04}{\sqrt{g_x}}\\\\v = 23.39 \sqrt{g_x}[/tex]
More about the kinematics link is given below.
https://brainly.com/question/25638908
