Answer:
4m/s
Explanation:
The motion is in downward direction, and the boat is moving with constant velocity, the position of the boat horizontally is 12m before the key fall.
The height difference "h"= ( y- y°)= 45 m
We can determine the needed time the key requires to reach the water using the expression from Newton's law
h= v(o) + 1/2 at^2
g= acceleration due to gravity= 9.8m/s^2
h= v(o) + 1/2 gt^2 -------------------------eqn*)
V(o)= 0 for the key= initial velocity
h= height that the key falls= 45m
If we substitute the above values into eqn(*) we have
We can make t^2 subject of the formula since v(0)= 0
t^2= h/(1/2 g)
= 45/(1/2× 9.8)
t^2= 45/4.9=9.18
t= √9.18
t=3.03 secs
Since the boat was moving with constant velocity,
Then Velocity= distance/ time
= 12/3.03
Velocity= 4m/s
Hence, the speed of the boat is 4m/s
Answer:
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.
