Question

chapter 2 linear motion problems a student launches an arrow upward with an unknown initial velocity. the arrow takes 2.3 seconds to reach its maximum height above the launch point. what was the initial velocity and height of the arrow

Answer 1

`V_(0)= 22.5(m)/(s)` and Ymax=25.8m

Step-by-step explanation:

Velocity at any time is given by
`V=V_(0)sin\theta -gt`. but when the arrow is on the top its velocity is zero and if it is launched upward the angle is 90°, so.

`0=V_(0) sin90-gt`

`V_(0)=gt=9.8(m)/(s^(2)).2.3s=22.5(m)/(s)`

At the maximun height, position is given by
`Ymax=V_(0)sin\theta.t-(1)/(2)gt^(2)`, replacing
`Ymax=22.5(m)/(s)x2.3s-(1)/(2)x9.8x(2.3)^(2)=25.8(m)/(s)`