A freely falling body covers two successive one-second intervals in distances whose ratio is 3:5. Taking g=10 m s⁻², the speed of the body at the beginning of the first of these two intervals is:
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A freely falling body covers two successive one-second intervals in distances whose ratio is 3:5. Taking g=10 m s⁻², the speed of the body at the beginning of the first of these two intervals is:

G1 Physics · Accepted Answer

Let the speed at the beginning of the first interval be v. Distance in first second: s₁ = v + 5. Speed at the beginning of the next second is v+10, so s₂ = (v+10)+5 = v+15. Given s₁:s₂ = 3:5, (v+5)/(v+15)=3/5. Therefore 5v+25=3v+45 ⇒ v=10 m s⁻¹.