Aball is thrown horizontally from the top of
a building 80 m high. the ball strikes the
ground 62 m horizontally from the point of
release.
what is the speed of the ball just before it
strikes the ground?
answer in units of m/s.

42.5 m/s

Explanation:

Given:

x₀ = 0 m

x = 62 m

y₀ = 80 m

y = 0 m

v₀ᵧ = 0 m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

Find: v

First, find the time it takes to land.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (80 m) + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 4.04 s

Find the horizontal component vₓ:

x = x₀ + vₓ t − ½ aₓ t²

(62 m) = (0 m) + vₓ (4.04 s) − ½ (0 m/s²) (4.04 s)²

vₓ = 15.3 m/s

Find the vertical component vᵧ:

vᵧ = aᵧ t + v₀ᵧ

vᵧ = (-9.8 m/s²) (4.04 s) + (0 m/s)

vᵧ = -39.6 m/s

Find the speed using Pythagorean theorem:

v = √(vₓ² + vᵧ²)

v = √((15.3 m/s)² + (-39.6)²)

v = 42.5 m/s

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i need the answer to this too but idk it ; -;