ASAP please help me

1.2 s

Explanation:

We'll begin by calculating the length (i.e distance) of the ramp. This can be obtained by using pythagoras theory as illustrated below:

NOTE: Length of the ramp is the Hypothenus i.e the longest side.

Let the Lenght of the ramp be 's'. The value of x can be obtained as follow:

s² = 4² + 3²

s² = 16 + 9

s² = 25

Take the square root of both side

s = √25

s = 5 m

Thus the length of the ramp is 5 m

Next, we shall determine the final velocity of the ball. This can be obtained as follow:

Initial velocity (u) = 3 m/s

Acceleration (a) = 2 m/s²

Distance (s) = 5 m

Final velocity (v) =?

v² = u² + 2as

v² = 3² + (2 × 2 × 5)

v² = 9 + 20

v² = 29

Take the square root of both side

v = √29

v = 5.39 m/s

Finally, we shall determine the time taken for the ball to reach the final position. This can be obtained as follow:

Initial velocity (u) = 3 m/s

Acceleration (a) = 2 m/s²

Final velocity (v) = 5.39 m/s

Time (t) =?

v = u + at

5.39 = 3 + 2t

Collect like terms

5.39 – 3 = 2t

2.39 = 2t

Divide both side by 2

t = 2.39 / 2

t = 1.2 s

Thus, it will take 1.2 s for the ball to get to the final position.

im guessing the west of station b

b. a lightning bug, plug of grass, and a worm