6. One can measure the gravitational acceleration by doing the following two mea

Question

6. One can measure the gravitational acceleration by doing the following two measurements. When throwing an object vertically upward, measure the time, A, it takes for it to travel upwards from a given height, y, and back to that same height, as well as the time. A12 , to travel from a height y2 and back. st (a) Work out an expression for the gravitational acceleration depending on (b) If the measurements are ,-527sA,-459s, and yry,-1m, what is the gravitational acceleration on the surface of that particular object? What might his object be?

Explanation / Answer

6. given delT1 is time taken to go from a height y1 to the same height

delT2 is time to travel from height y2 and back

now

let initial speed be v

then at height y

2*g*y = v^2 - u^2

u = sqrt(v^2 - 2gy)

hence

0 = u*delT - 0.5g*delT^2

u = 0.5g*delT

hence

sqrt(v^2 - 2gy1) = 0.5*g*delT1

sqrt(v^2 - 2gy2) = 0.5*g*delT2

(0.5g*delT1)^2 + 2gy1 = (0.5*delT2)^2 + 2gy2

g = (delT2^2 - delT1^2)/8(y1 - y2)

b. delt1 = 5.27 s

delT2 = 4.69 s

y2 - y1 = 1 m

then

g = 0.7221 m/s/s = 0.073608562 g = g/13.5

this is some moon of a planet like jupiter or saturn ( or some asteroid)

Related Questions

Navigate

• Browse (All)
• Browse 6
• Subjects

---

---

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