An arrow shot straight upwards reaches a maximum height of 104.0 m from the bow. (a) At what original speed in m/s does it leave the bow and (b) how long in seconds is the arrow in flight until it returns to the ground?
My answers came out to:
(a) 45.1 m/s
(b) 4.59 s. Or, should this be 4.59s * 2?
Thanks.
2007-09-12 06:46:04 · 3 answers · asked by labelapark 6 in Science & Mathematics ➔ Physics
t(up) = (2H/g)^0.5
...=(2*104.0/9.807)^0.5
...= 4.605 sec
Vf = Vi - gt, Vf = 0...
Vi = gt = 9.807 * 4.607
... = 45.16 m/sec
The arrow will be in flight for (assuming it was shot from exactly ground level):
t = t(up) + t(down)
...= 4.605 * 2 = 9.210 sec
All answers to 4 sig figs....I know this is a lot of precision, but the height was specified as 104.0 so that is the precision specified by the problem.
2007-09-12 06:58:16 · answer #1 · answered by gebobs 6 · 0⤊ 0⤋
It should be twice the time taken to travel one way.
Because the question mentions about the flight until it returns to ground.
104 = 0.5*9.8*t^2
t = 4.6s
v = u + at
=9.8*4.6
=45.08
2007-09-12 06:58:31 · answer #2 · answered by Savvy 2 · 0⤊ 0⤋
my head hurts.
2007-09-12 06:50:28 · answer #3 · answered by Anonymous · 0⤊ 0⤋