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grooveactiva
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Homework Statement
This is a physics problem that I made up in order to understand one aspect of kinematics in 1-dimension: A baseball is thrown straight up, leaving the hand which is 1.1 meter above the ground. It reaches a height of 12 meters. (a) What must be the speed of the baseball when it leaves the hand to reach this height? (b) How long does it take for the baseball to fall to the ground? (c) At what speed does it have just before it strikes the ground?
Homework Equations
y = yi + vi(Δt) + ½a(Δt)2
vf - vi = a(Δt)
vf2 - vi2 = 2a(Δs)
a =g = 9.8 m/s2
The Attempt at a Solution
(a) How fast baseball leaves hand:
vf2 - vi2 = 2(Δt)
vi = 0 m/s
vf2 + 0 = 2(-9.8 m/s2)(12-1.1 m)
vf = √(2*-9.8*10.9) = 14.6 m/s
(b) How long (how much time) does it take for baseball to fall to the ground:
vf - vi = a(Δt)
vf = 0 m/s because ball hits ground
(Δt) = (0 - 14.6 m/s)/-9.8 m/s2 = 1.5 seconds
Or, should I have used y = yi + vi(Δt) + ½a(Δt)2, with vf = 14.6 m/s
But then I get via the quadratic formula: time = -0.0074 s (discard because negative time) or 3.05 seconds
Why doesn't these two equations match in their anwers for time: 1.5 seconds vs. 3.05 seconds?
(c) At what speed does the baseball have just before it strikes the ground: Not sure how to proceed until I can figure out the answer to part (b) above.
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