A stone thrown vertically upward at12m/s. How high would the stone be when its speed is 8m/s?a.4.1mb.4.5mc.4.8md.4.9m
Question
A stone thrown vertically upward at12m/s. How high would the stone be when its speed is 8m/s?a.4.1mb.4.5mc.4.8md.4.9m
Solution
The height of the stone can be found using the equation of motion:
v^2 = u^2 + 2gs
where: v = final velocity u = initial velocity g = acceleration due to gravity (approximately -9.8 m/s², negative because it's acting downwards) s = displacement (height in this case)
Rearranging the equation to solve for s, we get:
s = (v^2 - u^2) / 2g
Plugging in the given values:
s = ((8 m/s)^2 - (12 m/s)^2) / (2*-9.8 m/s²) s = (64 m²/s² - 144 m²/s²) / -19.6 m/s² s = -80 m²/s² / -19.6 m/s² s = 4.08 m
Therefore, the stone would be approximately 4.1 m high when its speed is 8 m/s.
The correct answer is:
a. 4.1m
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