Question

A soccer ball is released from rest at the top of a grassy incline. After 4.4 seconds, the ball travels 92 meters and 1.0 s after this, the ball reaches the bottom of the incline. What was the magnitude of the ball's acceleration, assume it to be constant? Express your answer using two significant figures. a = nothing m/s2 Request Answer Part B How long was the incline? Express your answer using two significant figures. l = nothing m

Answer 1

The acceleration is 9.5 m / s

Step-by-step explanation:

This problem can be handled as a one-dimensional case, if we place a coordinate system along the gram, in this case with the initial data we can calculate the acceleration; and the ball leaves the rest its initial speed is zero

y = Vo t + ½ to t2

y = ½ to t2

a = 2 y / t2

a = 2 92 /4.42

a = 9.5 m / s2

The time or total to reach the bottom of the plane is

t = 4.4 + 1 = 5.4 s

We calculate the distance traveled

y= Vo t + ½ a t²

y = ½ 9.5 5.5²

y = 143 m = 14 x 101 m

Answer 2

A) The constant acceleration of the ball along the plane of the surface of the grassy incline = 9.50 m/s²

B) Length of the incline = 138.51 m

Step-by-step explanation:

A) The ball's acceleration is calculated with respect to the inclined plane; that is, the component parallel to the plane of the inclined plane.

u = 0 m/s (the ball is released from rest)

At t = 4.4 s,

x = 92 m

a = constant acceleration of the ball = ?

Using one of the equations of motion

x = ut + (1/2)at²

92 = (0)(4.4) + (1/2)(a)(4.4²)

92 = 9.68a

a = (92/9.68) = 9.50 m/s²

B) Length of the incline

u = 0 m/s

t = total time taken to reach the base of the incline = 4.4 + 1.0 = 5.4 s

a = constant acceleration of the ball along the plane of the surface of the incline = 9.50 m/s²

x = total distance along the surface of the incline travelled at the end of the 5.4 s = length of the incline = ?

Using one of the equations of motion again

x = ut + (1/2)at²

x = (0)(5.4) + (1/2)(9.5)(5.4²)

x = 138.51 m

Hope this Helps!!!