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? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible

Answer 1

Hello
We know that when the fisherman casts his line, it will travel in a parabolic path, typical to that of a projectile's motion.
Whenever an object is in projectile motion, its velocity has two components, the x-component and y-component. The x component is usually constant with acceleration 0; however, as we know that the acceleration due to gravity acts on any object moving vertically. This will be our acceleration for the vertical component. Now, we apply the equation of motion for the vertical component of velocity:
v^2 = u^2 +2as; where v is the y-component final velocity, u is y-component of the initial velocity, s is the distance traveled and a is the acceleration.
Whenever an object attains "maximum height", its vertical velocity instantaneously becomes 0; therefore, v = 0.
a = -9.81 m/s^2 (negative sign because velocity is upwards while acceleration downwards)
s = 2.9 m as stated
Plugging the values into the equation, u (vertical component of initial velocity) works out to be:
7.54 m/s in the vertical direction.
The vertical component of the velocity is:
velocity * sin(angle with horizontal)
Thus, velocity =
7.54/sin(25)
= 17.8 m/s