A baseball 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.

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

AkshayG AkshayG · Answer 2 · 2020-01-14T10:58:08+01:00

(a) The expression for the initial horizontal velocity of the ball is [tex]\boxed{v_{0x}=v_f\,\text{cos}\,\theta_f}[/tex].

(b) The initial vertical velocity of the ball is [tex]\boxed{17.39\,\text{m/s}}[/tex].

(c) The magnitude of the ball’s initial velocity is [tex]\boxed{34\,\text{m/s}}[/tex].

(d) The angle above the horizontal at which the ball left is [tex]\boxed{30.74^\circ}[/tex].

Further Explanation:

Given:

The velocity at which the ball is caught by the fan is [tex]32.4\,\text{m/s}[/tex].

The angle made by the ball as it reaches the fan is [tex]25.5^\circ[/tex] below the horizontal.

The height at which the ball was caught by the fan is [tex]5.5\,\text{m}[/tex].

Concept:

Part (a):

The ball moves in two-dimensional motion. The vertical component of the ball changes due to the effect of acceleration due to gravity in the vertical direction. But the horizontal component of the velocity of ball always remains the same as there is no acceleration in horizontal direction.

Therefore, the horizontal component of the initial velocity of the ball can be obtained from the final velocity and the angle made by the velocity with the horizontal.

[tex]\boxed{v_{0x}=v_f\,\text{cos}\,\theta_f}[/tex]

Here [tex]v_{0x}[/tex] is the initial horizontal velocity of the ball, [tex]v_{f}[/tex] is the final velocity of the ball and [tex]\theta_{f}[/tex] is the final angle made by the ball.

Part (b):

The initial vertical velocity of the ball can be obtained from the newton’s third equation of motion by establishing the relation between the initial and final velocity of the ball along with the height at which the ball was caught.

The final vertical velocity of the ball at the moment when it was caught is given by.

[tex]v_{fy}=v_f\,\text{sin}\,\theta_f[/tex]

Here, [tex]v_{fy}[/tex] is the final velocity in the vertical direction.

Substitute [tex]32.4\,\text{m/s}[/tex] for [tex]v_f[/tex] and [tex]25.5^\circ[/tex] for [tex]\theta_f[/tex] in above expression.

[tex]\begin{aligned}v_{fy}&=(32.4\,\text{m/s})\,\text{sin}(25.5^\circ)\\&=13.95\,\text{m/s}\end{aligned}[/tex]

The initial vertical velocity of the ball is obtained by.

[tex]v^{2}_{fy}=v^{2}_{0y}-2\,\text{as}[/tex]

Substitute [tex]13.95\,\text{m/s}[/tex] for [tex]v_{fy}[/tex], [tex]9.8\,\text{m/s}^2[/tex] for [tex]g[/tex] and [tex]5.5\,\text{m}[/tex] for [tex]s[/tex] in above equation.

[tex]\begin{aligned}(13.95)^2&=v^{2}_{0y}-2\times9.8\times5.5\\v_{0y}&=\sqrt {194.60 + 107.8} \\&=17.39\,\text{m/s}\end{aligned}[/tex]

Thus, the initial vertical velocity of the ball is [tex]\boxed{17.39\,\text{m/s}}[/tex].

Part (c):

The magnitude of the initial velocity of the ball is given by:

[tex]v_0=\sqrt{v^{2}_{0x}+v^{2}_{0y}}[/tex]

Here, [tex]v_{0}[/tex] is the initial velocity of the ball.

Substitute the values in above expression.

[tex]\begin{aligned}v_{0}&=\sqrt{(v_f\,\text{cos}\,\theta_f)^2+(17.39)^2}\\&=\sqrt{(32.4\,\text{cos\,(25.5}^\circ))^2+302.41}\\&=\sqrt{1157.60}\\&\approx 34\,\text{m/s}\end{aligned}[/tex]

Thus, the magnitude of the initial velocity of the ball is [tex]\boxed{34\,\text{m/s}}[/tex].

Part (d):

The angle made by the ball with the horizontal is given by:

[tex]\theta=\text{tan}^{-1}\left( {\dfrac{{{v_{0y}}}}{{{v_{0x}}}}} \right)[/tex]

Here, [tex]\theta[/tex] is the angle made by the ball as it is hit by the bat.

Substitute the values of [tex]v_{0y}[/tex] and [tex]v_{0x}[/tex] in above expression.

[tex]\begin{aligned}\theta&=\text{tan}^{-1}\left(\dfrac{17.39}{32.4\,\text{cos}\,25.5^\circ}\right)\\&=\text{tan}^{-1}(0.595)\\&=30.74^\circ\end{aligned}[/tex]

Thus, the angle made by the ball with the horizontal is [tex]\boxed{30.74^\circ}[/tex].

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Answer Details:

Grade: College

Chapter: Motion in one dimension

Subject: Physics

Keywords: Baseball, slugger, watches the ball, visiting with the fan, caught, angle, final velocity, horizontal, component, third equation of motion.