. A boy is sitting in his red wagon at the top of a hill, holding onto a tree to keep from rolling down the hill. The distance from the top of the hill to the bottom is 41 meters. He pushes off from the tree at a starting velocity of 2 m/s, and then gravity accelerates the wagon at a constant rate of 2 m/s2. How long will it take him to reach the bottom of the hill (round to the nearest hundredth)?
h(t) = 41 -2t - t^2
h(t) = 0 when t = √42 - 1 = 5.48
To find the time it takes for the boy to reach the bottom of the hill, we can use the equations of motion and the given information.
The initial velocity (u) of the wagon is 2 m/s, and the acceleration (a) due to gravity is 2 m/s^2. The distance traveled (s) is 41 meters.
We can use the second equation of motion: s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the boy starts from rest at the top of the hill, his initial velocity (u) is 0 m/s. Therefore, the equation becomes: s = (1/2)at^2.
Substituting the given values: 41 = (1/2)(2)t^2.
Simplifying the equation: 41 = t^2.
Taking the square root of both sides, we get: t = √41.
Calculating the square root of 41 using a calculator or any math tool, we find that t is approximately 6.4 (rounded to the nearest hundredth).
Therefore, it will take the boy approximately 6.4 seconds to reach the bottom of the hill.