I am confused on how to solve the following problem An arrow is shot vertically

Question

I am confused on how to solve the following problem An arrow is shot vertically upward. 3 seconds later, it is ata height of 35cm. What was the inital speed of the arrow? how longis the arrow in flight from launch to returning to the intialheight? (assume arrow falls verticlly downward) Any help would be greatly appreciated!!! I am confused on how to solve the following problem An arrow is shot vertically upward. 3 seconds later, it is ata height of 35cm. What was the inital speed of the arrow? how longis the arrow in flight from launch to returning to the intialheight? (assume arrow falls verticlly downward) Any help would be greatly appreciated!!!

Explanation / Answer

1) the acceleration of the arrow on the string is not part ofthis problem. t=0 is when the arrow leaves the string at astarting velocity Vo. 2) the height at this time is 0
The equation of interest is for this is
S = 1/2 at2 + Vot + So Inserting knownvalues, we get
.35 = (-4.9)9 + 3Vo Note that i put s inmeters
3Vo = 44.1 + .35 ; Vo = 14.817 m/sec
It's time to think for a second rather than jumping to newequations. If the initial velocity is almost 15 m/sec, thearrow will go through .35 meters the first time very quickly. That means that the .35 meters in this problem is on the waydown, and we can expect the arrow to hit the ground very soon, asthe arrow will be going almost 15 m/sec down at this point.
S=0 twice, at launch and at return. We are interested inthe non-zero answer.
0 = 1/2 (-9.8) t2 + 14.817t Since tis not zero, we can divide both side by t to get
0 = -4.9t + 14.817; t = 14.817/4.9 = 3.024 seconds

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