Frank A. answered • 02/12/20
Tutor
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Knowledge, experience and insight from decades of teaching physics.
Jessica M. Here is how it works:
Mass is superfluous for this problem.
Initial vertical velocity Vi is 30sin28°
Final vertical velocity Vf = 0
Acceleration = g = -9.81ms-2
Since Vf = Vi + g•t, t = (Vf -Vi)/g
substituting
(0-14.08)/-9.81 = t = 1.4356s however this is the time to reach maximum height
∴ total time = 1.4356•2 = 2.87s or 3s (1 sig fig)
If you need further clarification on this or any other problem, just ask.
Mr. Frank A
Jessica M.
Thank you so much! If there is another problem that is exactly the same but just asking for meters instead of seconds do I still solve it that same way or take a different approach?
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02/12/20
Frank A.
tutor
It depends if the distance is height or range (horizontal). For vert. Use 2(-g)x=0-30sin28. If it's horizontal then use total time of flight times 30cos28. Hope that helps.
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02/12/20
Jessica M.
Thank you so much!02/12/20