A steel ball is dropped from a building\'s roof and passes a window,taking 0.127

Question

A steel ball is dropped from a building's roof and passes a window,taking 0.127s to fall from the top tothe bottom of the window, a distance of 1.16 m. It then falls to asidewalk and bounces back past the window, moving from the bottomto top in 0.127s. assume that the upward flight is an exact reverseof the fall. The time spent below the bottom of the window is2.02s. How tall is the building?
I did this 4 times and got 4 different answers of 34.69, 24.6,44.6, and 34.7. I'm not sure, but my h1=1.16, h2=3.6963, andh3=different answers ie: 29.83, 39....
I did this 4 times and got 4 different answers of 34.69, 24.6,44.6, and 34.7. I'm not sure, but my h1=1.16, h2=3.6963, andh3=different answers ie: 29.83, 39....

Explanation / Answer

you just need to use the three eqns of motion let speed at top of window be u, speed on reaching itsbottom be v. So, v = u + gt => v = u + 9.8*0.127 m/s And, v*v = u*u + 2gh => (u+9.8*0.127)2 - u2 =2*9.8*1.16 => 2*9.8*0.127*u + 0.127*0.127*9.8*9.8 =2*9.8*1.16 => u = 8.512 m/s => v = 9.756 m/s => speed at bottom of building being V, V = v + g*t', t' being 2.02/2 = 1.01s since the path issymmetric => V = 9.756 + 9.8*1.01 m/s = 19.654 m/s So, speed at top of building being U=0 m/s, V*V = U*U + 2gH H being the height of the building. => H = 19.654*19.654/19.6 m = 19.708 m ~ 19.71 m

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