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Help with this physics problem?!?
Please help!
Ball A is dropped from a height h above the ground. At precisely half the time that it takes ball A to reach the ground, a second ball, ball B, is thrown downward from the same height. Both balls hit the ground simultaneously.
Show that the speed with which ball B is thrown downward is given by the expression Vb= (3/4)*sqrt(2gh).
Please help my final exam is tomorrow and I must know how to do this.
1 Answer
- Seamus OLv 78 years agoFavorite Answer
Take down as the positive direction
For ball A:
v(i) = 0 m/s ... [just dropped]
a = g
s = h
s = v(i)t + (1/2)at²
h = 0 + (1/2)g t(a)²
h = (g/2)t(a)² .......................... [eqn 1]
also v(f)² = v(i)² + 2as
v(f)² = 0 + 2gh
v(f) = √(2gh)
v(f) = v(i) + at
√(2gh) = 0 + gt(a)
t(a)= √(2gh) / g ................................... [eqn 2]
For ball B:
v(i) = ?
t(b) = t(a) / 2
a = g
s = h
s = v(i)t + (1/2)at²
h = v(i) * [(t(a) / 2] + (g/2) * [t(a)² / 4]
h = v(i) * [(t(a) / 2] + [gt(a)² / 8] ................................... [eqn 3]
Now [eqn 1] = [eqn 2]
so (g/2)t(a)² = v(i) * [(t(a) / 2] + [gt(a)² / 8]
4gt(a)²/8 = v(i) * [(t(a) / 2] + [gt(a)² / 8]
3gt(a)² / 8 = v(i) * [(t(a) / 2]
3gt(a)² / 4 = v(i) * t(a)
v(i) = 3gt(a) / 4
Now subs [eqn 2] → v(i) = (3g / 4) * [√(2gh) / g]
v(i) = (3/4) √(2gh)
So ball B is thrown downward with a speed of (3/4) √(2gh)
