A basketball is thrown horizontally with an initial speed of
4.50 m/s (Figure 1). A straight line drawn from the
release point to the landing point makes an angle of 30.0
º with the horizontal. What was the release height?

1.35 m

Explanation:

Taking down to be positive, given:

Δx = Δy / tan 30.0º

v₀ₓ = 4.50 m/s

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = 10 m/s²

Find: Δy

First, find the time it takes to land in terms of Δy.

Δy = v₀ t + ½ at²

Δy = (0 m/s) t + ½ (10 m/s) t²

Δy = 5t²

Next, find Δx in terms of t.

Δx = v₀ t + ½ at²

Δx = (4.50 m/s) t + ½ (0 m/s) t²

Δx = 4.50t

Substitute:

Δy = 5 (Δx / 4.50)²

20.25 Δy = 5 (Δx)²

4.05 Δy = (Δx)²

4.05 Δy = (Δy / tan 30.0º)²

4.05 Δy = 3 (Δy)²

1.35 = Δy

The basketball was thrown from an initial height of 1.35 m.

Graph: desmos.com/calculator/ujuzdo9xpr

the answer would be c. motion and direction

i hope this : )