Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10 m/s. How long after Bill released his rock should Ted throw his if they want the stones to hit the water simultaneously? Choose answer closest to calculated result. A. del t 15.5 sec B. del t 0.86 sec C. can not be determined. D. del t - 0.72 sec

D) - 0.72 secs

Explanation:

Parameters given:

Height of bridge = 40ft = 12.19 m

Initial velocity of Bill's stone = 0m/s

Initial velocity of Ted's stone = 10m/s

We find the time it take Bill's stone to bit the river and the time it takes Ted's stone to hit the river. Then we find the time difference.

Using one of the equations of motion:

For Bill:

S = ut + ½gt²

Where g = 9.8 m/s

12.19 = 0 + ½*9.8*t²

t² = 12.19/4.9 = 2.49

t = 1.58 secs

For Ted:

S = uT + ½gT²

12.19 = 10*T + ½*9.8*T²

=> 4.9T² + 10T - 12.19 = 0

Using quadratic formula and retaining only the positive value, we get that:

T = 0.86 secs

Time difference between Bill's throw and Ted's throw is:

0.86 - 1.58 = - 0.72 secs

In reality, this means that Ted must throw his stone 0.72 secs before Bill throws his for both stones to land the same time.

50 inches should be the resultant vector