A key falls from a bridge that is 45 m above the water. It falls directly into a model boat, moving with constant velocity, that is 12 m from the point of impact when the key is re- leased. What is the speed of the boat

The person above me is correct. The speed of the boat is 3.6363m/s, or also 4m/s.

We know these following values from the key:

d = 45 , g = -9.8 (its because its falling) , vi = 0

With this, we can plug and chug using kinematic equations.

Doing so will get you 29.69 for Vf and 3.03 for time.

Now, we want the speed from the boat -- speed is defined as the distance over time. We know the distance from the boat is 12m, but now we know the time, which is 3.03.

-This time is correct because we know the boat was at "0m" when the key was at the start, but after it dropped, it was at 12m. So by finding out the time it took for the key to drop, it also tells us the amount of time it took the boat to travel 12m.

So this means we can divide 12m by 3.03s, or you can round it to 3. Whatever really works;

12/3.03 should get you something like 3.63 repeating

if you rounded, you will get 12/3, which is 4.

the maxium height is 34m/s

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft