Question

chegg a projectile is shot directly away from earth's surface. neglect the rotation of the earth. what multiple of earth's radius gives the radial distance the projectile reaches if its initial speed is 0.745 of the escape

Answer 1

The radial distance of the projectile reaches  reaches if its initial speed is 0.745 of the escape is (25/24) R.

Calculation:-

The mass of the earth is M and radius of the earth is R.

The escape velocity ve = √(2GM/R)

Since the terminal veocity v0 = (1/5) ve  = (1/5) √(2GM/R)

(a) Let h be the height from the surface of the earth

From the law of conservation of energy

(KE)1 = Change in potential energy

(1/2) m (v0)2 = (GMm/R) - (GMm/(R+H))

(1/2) m (1/25) (2GM/R) = (GMm/R) - (GMm/(R+H))

(1/25R) = (1/R) - (1/(R+H))

1/25 = H / R+H

R+H = 25H

H = R/24

Now R+H = (25/24) R

velocity can be the perception of the fee at which an object covers distance. A quick-transferring item has a excessive pace and covers a substantially large distance in a given amount of time, whilst a sluggish-transferring item covers a incredibly small quantity of distance in the identical quantity of time.

The primary unit or the S.I. unit of velocity is m/s or ms⁻¹.

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