The distance between the two vessels when they land below the waterfall is approximately 10.65 meters
The reason why the above value is correct is given as follows;
The given parameters of the motion are;
The height of the water fall, h = 4.6 m
The speed of the spy's boat, vââ = 17 m/s
Speed of the official's boat, vââ = 28 m/s
Acceleration of gravity, g = 9.81 m/s²
Required:
The difference between the distance travelled by the boats when they land below the waterfall
Solution:
The time it takes the boats to reach the bottom of the waterfall, t, is given as follows;
[tex]h = u\cdot t + \dfrac{1}{2} \cdot g \cdot t^2[/tex]
Where the vertical component of velocity is zero for free fall , we have;
u = 0 m/s
[tex]t = \sqrt{\dfrac{2 \cdot h}{g} }[/tex]
Therefore, we have;
[tex]t = \sqrt{\dfrac{2 \times 4.6}{9.81} } \approx 0.9684[/tex]
The time it takes both the spy and the official boat to land, t â 0.9684 s
The difference in the horizontal distance travelled by the two boats when they land, d, is given as follows;
Projectile motion xy equation
d = t Ă (vââ - vââ)
â´ d â 0.9684 s Ă (28 m/s â 17 m/s) = 10.6524 m â 10.65 m
The distance apart they will be when they land below the waterfall, d â 10.65 meters
Learn more about free fall motion here:
https://brainly.com/question/17207644
