Asked by Anonymous
A baseball is thrown from the roof of 23.5m -tall building with an initial velocity of magnitude 10.0m/s and directed at an angle of 53.1∘ above the horizontal.
a.What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.
b.What is the answer for part (A) if the initial velocity is at an angle of 53.1∘ below the horizontal?
a.What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.
b.What is the answer for part (A) if the initial velocity is at an angle of 53.1∘ below the horizontal?
Answers
Answered by
Steve
well, you have
dx/dt = 10cos53.1° = 6
dy/dt = 10sin53.1° - 9.8t = 8-9.8t
y = 23.5+8t-4.9t^2
y=0 when t=3.15
At t=3.15,
dx/dt = 6
dy/dt = -22.87
so, the speed is √(6^2+22.87^2) = 23.64 m/s
Given that, I expect you can do part (b).
dx/dt = 10cos53.1° = 6
dy/dt = 10sin53.1° - 9.8t = 8-9.8t
y = 23.5+8t-4.9t^2
y=0 when t=3.15
At t=3.15,
dx/dt = 6
dy/dt = -22.87
so, the speed is √(6^2+22.87^2) = 23.64 m/s
Given that, I expect you can do part (b).
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