Question

Problem 7: A baseball slugger hits a pitch and watches the ball fly into the bleachers for a home run, landing h = 9.5 m higher than it was struck. When visiting with the fan that caught the ball, he learned the ball was moving with final velocity vf = 39.6 m/s at an angle θf = 33° below horizontal when caught. Assume the ball encountered no air resistance, and use a Cartesian coordinate system with the origin located at the ball's initial position.

(a) Calculate the ball’s initial vertical velocity, v0y, in m/s.
(b) Calculate the magnitude of the ball’s initial velocity, v0, in m/s.
(c) Find the angle θ0 in degrees above the horizontal at which the ball left the bat.

Answer 1

Answer:

Part a)

$v_y = 25.52 m/s$

Part b)

$v = 41.9 m/s$

Part c)

$\theta = 37.5 degree$

Explanation:

Part a)

As we know that final velocity is

$v_f = 39.6 m/s$

angle made by it is given as

$\theta_f = 33^o$

now we know its two components are

$v_{fy} = 39.6 sin33 = 21.56 m/s$

$v_{fx} = 39.6 cos33 = 33.21 m/s$

now we can use kinematics in Y direction

$v_f^2 - v_i^2 = 2 a d$

$21.56^2 - v_y^2 = 2(-9.81)(9.5)$

$v_y = 25.52 m/s$

Part b)

Also we know that velocity in x direction will remains same

so

$v_x = 33.21 m/s$

so net speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{33.21^2 + 25.52^2}$

$v = 41.9 m/s$

Part c)

Angle of projection is given as

$tan\theta = \frac{v_y}{v_x}$

$tan\theta = \frac{25.52}{33.21}$

$\theta = 37.5 degree$