A hopper jumps straight up to a height of 1.3 m. With what velocity did it leave the floor?

Answer 1

Let s= v₀t -gt²/2 where g=9.8 m/sec² and v₀ is the initial velocity. The velocity at the maximum height is 0 so v= v₀ -gt =0 and v₀ =gt or t= v₀/g for the time to reach 1.3m So the equation becomes 1.3 = v₀(v₀/g) -4.9(v₀²/g²) or 1.3 = v₀²/9.8 -v₀²(4.9/9.8²) = v₀²/9.8( 1-1/4.9•9.8) and so 1.3•9.8 = v₀²(1-1/4.9) = v₀²(3.9/4.9). This gives v₀²= 16 and v₀ =4 m/sec.