Abaseball is thrown straight out from a mound at 40 m/s and lands 21m away. how tall is the mound

1.35 m

Explanation:

First, find the time it takes to land.  Given in the x direction:

Δx = 21 m

v₀ = 40 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

(21 m) = (40 m/s) t + ½ (0 m/s²) t²

t = 0.525 s

Next, find the height.  Given in the y direction:

v₀ = 0 m/s

a = 9.8 m/s²

t = 0.525 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.525 s) + ½ (9.8 m/s²) (0.525 s)²

Δy = 1.35 m

Round as needed.