A spaceship hovering over the surface of venus drops an object from a height of

Question

A spaceship hovering over the surface of venus drops an object from a height of 51 m. how mich longer does it take to reach the surface than if a dropped from the same height on earth? neglect air resistance in both cases. (the acceleration due to gravity on venus is 90.7% of that on earth. gvenus= (0.907)g ] A spaceship hovering over the surface of venus drops an object from a height of 51 m. how mich longer does it take to reach the surface than if a dropped from the same height on earth? neglect air resistance in both cases. (the acceleration due to gravity on venus is 90.7% of that on earth. gvenus= (0.907)g ]

Explanation / Answer

Time taken on venus will be given by:

Hv = V0*tv + 0.5*gv*tv^2

V0 = 0 m/sec

Hv = 51 m

gv = 0.907*9.81 = 8.897 m/sec^2

51 = 0*tv + 0.5*8.897*tv^2

tv = sqrt (51*2/8.897) = 3.386 sec

Nowtime taken on earth will be

te = sqrt (51*2/9.81) = 3.224

So

dt = tv - te = 3.386 - 3.224 = 0.162 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.