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

Answer:

b) 18 m/s

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 $v_{y}=0$

$0=v_{oy} ^{2} +2gy\\0=v_{oy} ^{2} -2(9.8)(2.9)$

$v_{oy}=\sqrt{2(9.8)(2.9)} =7.53m/s$

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

$v_{oy}=v_{o}sin\beta$

$v_{o}=\frac{v_{oy} }{sin(25)} =\frac{7.53m/s}{sin(25)} =18m/s$

So, the answer to the problem is b