Abaseball slugger hits a pitch and watches the ball fly into the bleachers for a home run, landing h = 5.5 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 = 32.4 m/s at an angle θf = 25.5° 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) create an expression for the ball's initial horizontal velocity, v0x, in terms of the variables given in the problem statement. b) calculate the ball's initial vertical velocity, v0y, in m/s. c) calculate the magnitude of the ball's initial velocity, v0, in m/s. d) find the angle, theta0, in degrees above the horizontal at which which the ball left the bat.

a) create an expression for the ball's initial horizontal velocity, V0x, in terms of the variables given in the problem statement.

v0x = vf * cos(Θf)

 
b) calculate the ball's initial vertical velocity, V0y, in m/s

v0x = 32.4m/s * cos(-25.5º) = 29.2 m/s 
tanΘ = v1y / v0x → tan(-25.5) = v1y / 29.2m/s → v1y = -13.93 m/s 
the vertical velocity when the ball was caught. 

(v0y)² = (v1y)² + 2as = (-13.93m/s)² + 2 * 9.8m/s² * 5.5m = 301.78 m²/s² 
v0y = 17.37 m/s 

c) calculate the magnitude of the ball's initial velocity, v0, in m/s

v0 = sqrt (v0y^2 + v0x^2)

v0 = sqrt (17.37^2 + 29.2^2) m/s

v0 = 33.98 m/s 

d) find the angle, theta0, in degrees above the horizontal at which which the ball left the bat.

tan Θ = v0y/v0x

Θ = arctan(17.37/29.2) = 30.75º above horizontal

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explanation:

i think the answer is a and b

i wish you luck