An object is dropped from a bridge a second object is thrown down one second later they both reach the water twenty meters below what was the initial speed of the second object?

Answer 1

The question seems ambiguous. Are you saying that both objects reach the water exactly one second after the first object is dropped? Are you saying that the second object was thrown exactly one second after the first object was dropped? I'm assuming the idea is that the objects reach the water at the same moment, and that you are ignoring the effect of air friction on the objects. A2 I assume object 2 is thrown 1 sec after object 1 is dropped and they both reach the bottom at the same time. Taking down as positive, the equation for each object is: L = (1/2)g(T - t1)^2 L= V(T-t2) + (1/2)g(T-t2)^2 Where T is the time they reach bottom. t1 & t2 are the initial times each object leaves. Choose t1 = 0 , then t2 = 1 (given). Also L=20 (given). And V is the initial velocity of object 2. This can be worked algebraically by solving for T in eq1 and substituting in eq2. Then solve for V. However, if you make the approximation g=10 (instead of 9.8) then it can be solved almost by inspection, without a calculater. Putting in the known values gives; 20 = 5T^2 20 = V(T-1) + 5(T-1)^2 From eq1, T=2 , so eq2 becomes; 20 = V + 5 , and so V = 15 which should be pretty close to the correct answer.