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 $2.9 \mathrm {~m}$ What was the initial velocity he launched the bait with? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible.
A)18 m/s
B)7.9 m/s
C)7.6 m/s
D)6.3 m/s

Question

Quizplus · Accepted Answer

The correct answer is A because the initial velocity is the velocity at which the bait and hook leave the fisherman's hand. The maximum height reached by the bait and hook is determined by the initial vertical velocity component, which is given by $v_i \sin \theta$. The maximum height is also determined by the acceleration due to gravity, which is $g = 9.8 \mathrm{~m/s^2}$. Using the equation of motion for vertical motion, we can find the initial vertical velocity component:$$v_{yf}^2 = v_{yi}^2 + 2g(y_f - y_i)$$$$0 = v_i^2 \sin^2 \theta - 2g(2.9 \mathrm{~m})$$$$v_i = \sqrt{\frac{2g(2.9 \mathrm{~m})}{\sin^2 \theta}}$$$$v_i = \sqrt{\frac{2(9.8 \mathrm{~m/s^2})(2.9 \mathrm{~m})}{\sin^2 25^\circ}}$$$$v_i = 18 \mathrm{~m/s}$$

A Fisherman Casts His Bait Toward the River at an Angle