Question

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its launch point.

A) What was the arrow's initial speed?
B) How long did it take for the arrow to first reach a height of 17.5 m
above its launch point?

Answer 1

A. 686 m/s B. 70 s

Step-by-step explanation:

A. Assuming we have a constant acceleration due to gravity of 9.8m/s^2, we can find our answer through the manipulation of the equation,

`v-f^2=v_i^2+2ad\\v_i = √(vf^2-2ad)\\v_i=√(0-2*-9.8*35) = 686m/s`

B. We can manipulate another kinematic equation to solve for time in this case

`v_f = v_i +a*t\\t=(v_f-v_i)/(a) =(0-686)/(-9.8) = 70s`