It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?
The general equations of accelerated motion apply to falling (or rising) bodies with the exception that the term "a" for acceleration is replaced by the term "g). This results in
....Vf = Vo + gt (the acceleration is assumed constant)
....d = Vo(t) + g(t^2)/2
....Vf^2 = Vo^2 + 2gd
As written, these expressions apply to falling bodies. The equations that apply to rising bodies are
....Vf = Vo - gt (the acceleration is assumed constant)
....d = Vo(t) - gt^2/2
....Vf^2 = Vo^2 - 2gd
From Vf = 0 = 100 - 9.8t, t(up) = 10.2sec. = t(dwn) making the total flight time 20.4sec.
From h = Vot - gt^2/2,
h = 100(10.2) - 9.8(10.2)^2/2 or
h = 510.2m.
It is possible to shoot an arrow at a speed as high as 100 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air?
Please help! I do not have any clue as to how to solve. Thanks!
3 answers
Thanks
I don't quite understand how you determined the time without it being given to you.
2 answers