Question

A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the bait and hook reach a maximum height of 2.9 m. What was the initial velocity he launched the bait with?

a) 6.3 m/s
b) 18 m/s
c) 7.6 m/s
d) 7.9 m/s

Answer 1

b) 18 m/s

Step-by-step explanation:

From the exercise we got maximum height and the angle which the bait is launched.

`\beta =25`º

`y=2.9m`

We know that at maximum height, the velocity at y-axis is 0

`v_(y) ^(2) =v_(oy) ^(2)+2a(y-y_(o))`

Since
`v_(y)=0`

`0=v_(oy) ^(2) +2gy\\0=v_(oy) ^(2) -2(9.8)(2.9)`

`v_(oy)=√(2(9.8)(2.9)) =7.53m/s`

Since the bait is launched at
`\beta =25º`

`v_(oy)=v_(o)sin\beta`

`v_(o)=(v_(oy) )/(sin(25)) =(7.53m/s)/(sin(25)) =18m/s`

So, the answer to the problem is b