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lakkis [162]
3 years ago
13

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 t

he 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.
How many seconds will it take the object to reach the ground and what will be its impact speed with the ground?

The ball will not drop until your submit your answers.
Physics
1 answer:
sattari [20]3 years ago
8 0

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

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Check attachment. For free body diagram and better understanding

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Explanation:

Given that,

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Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

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Put the value of \omega=3.14\ rad/s in equation (I) and (II)

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Put the value of \omega=9.42\ rad/s in equation (I) and (II)

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