A ball is thrown vertically upward from the top of a tower with an initial speed of 12 m/s. It reaches the ground 4.0 s later. What is the height of the tower? (Ignore air resistance and take g = 10 m/s².)
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Correct answer
A. 32 m
Principle or equation
For uniformly accelerated motion, displacement is given by s = v₀t + (1/2)at², with upward direction positive and a = -g.
Why this answer is correct
Take upward as positive. The displacement from the top to the ground is -h. Using s = v₀t + (1/2)at², with v₀ = 12 m/s, t = 4.0 s, a = -10 m/s²: -h = 12(4) + (1/2)(-10)(4²) = 48 - 80 = -32 m. Thus h = 32 m.
Example
If a ball is thrown upward at 10 m/s from a cliff and hits the ground after 3 s, then -h = 10(3) - 5(9) = 30 - 45 = -15 m, so h = 15 m.
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