Afisherman casts his bait toward the river at an angle of 25° above the horizontal. the bait and hook reach a maximum height of what was the initial velocity with which the bait was cast? assume that the fishing line exerts no appreciable drag force on the bait and hook.

the vertical component of the casting velocity will be reduce to zero at point h by the force of gravity.

If we use the kinematic equation

vf² = vi² +2as

and realize that the final vertical velocity is zero

and the initial vertical velocity is vsinθ

where θ is the angle of cast above the horizon

0 = vsin25 + 2(-9.8)h

h = 0.423v / 19.6

h = 0.02156v

so the missing potential heights and the associated answer velocities are

v = 19 m/s...h = 0.410 m

v = 8.3 m/s...h = 0.179 m

v = 6.9 m/s...h = 0.149 m

v = 8.4 m/s...h = 0.181 m

choose the one that fits the question

or use the method to find the correct velocity based on the actual height reference frame.

answer: 15

step-by-step explanation:

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.