Aboy standing on a 19.6 meter tall bridge sees a motorboat approaching the bridge at a constant speed. when the boat is 27 meters from the bridge, the boy drops a stone to the water below. if the stone strikes the water 3.0 meters in front of the boat, at what speed is the boat traveling?

A. 12 m/s

Explanation:

Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.

The equation that describes the motion of the stone is:

y = y_0 + v_0 * t+1/2 * a * t^2

The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:

-19,6 m = -1/2 * 9,8 m/s^2  * t^2

t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2

t=√(4,012 s^2) = 2,003 s

Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:

v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s

Have a nice day! :D

the answer is a. at the equator