- #1
jamiebean
- 55
- 3
- Homework Statement
- A boy throws a stone vertically up with a speed 24.6 m/s. Assuming that the stone's motion is affected by the gravitaional acceleration only (acceleration g=9.8ms-2). What is the displacement in m between the stone's position at time 3.8 s and its initial position? (hint: displacement has a sign "+" or "-", the canvas system can accept signed input. upward direction is "+" and the downward direction is "-", gravitation acceleration is negative in sign)
- Relevant Equations
- d=vi x t + (at^2)/2
gravitation acceleration= -9.8
I attempted the question with
d=vi x t + (at^2)/2
gravitation acceleration= -9.8
and I got the solution of 22.724.
Should I use the value of -9.8? or should I just use 9.8?
should I use the equation above? I feel like what I am calculating is not displacement but distance...
thank you
d=vi x t + (at^2)/2
gravitation acceleration= -9.8
and I got the solution of 22.724.
Should I use the value of -9.8? or should I just use 9.8?
should I use the equation above? I feel like what I am calculating is not displacement but distance...
thank you
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