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.

6^2 or 36

step-by-step explanation:

6000n

6×1000n

6×10^3 n

so we need 6n need to be a perfect cube

so n has to be 6^2 so that 6n is 6^3

what i'm saying is if you let n be 6^2 you have 6^3×10^3 or 60^3 which is a perfect cube