If a bowling ball ball was released from the 6th floor of a hotel room 22 meters from the ground, at what velocity would the bal
l be moving just before it hits the head of the man who is 1.5 meters tall? Solve using kinematic equations, unknown values, and SOHCAHTOA.
1 answer:
Explanation:
Given:
y₀ = 22 m
y = 1.5 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2 (-9.8 m/s²) (1.5 m − 22 m)
v = -20.0 m/s
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