A ball is thrown straight up and passes point B (at a height of 56.8 m above its starting point O) in 5.8 s. What was its initial speed? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s

38.2 m/s

Explanation:

Given:

Δy = 56.8 m

a = -9.8 m/s²

t = 5.8 s

Find: v₀

Δy = v₀ t + ½ at²

56.8 m = v₀ (5.8 s) + ½ (-9.8 m/s²) (5.8 s)²

v₀ ≈ 38.2 m/s

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.