A softball is hit over a third baseman's head with speed and at an angle theta from the horizontal. Immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity V = 7.00 m/s, for a time t = 2.00 s. He then catches the ball at the same height at which it left the bat. The third baseman was initially L = 18.0 m from the location where the ball was hit at home plate. Find v0, the initial speed. Use g = 9.80 m/s2 for the magnitude of the acceleration due to gravity. Express your answer numerically in meters per second to three significant figures. Find the angle theta in degrees. Express your answer numerically in degrees to three significant figures. Find a vector expression for the velocity of the softball 0.100 s before the ball is caught. Use the notation vx, vy, an ordered pair of values separated by a commas Express your answer numerically in meters per second to three significant figures. Find a vector expression for the position of the softball 0.100 s before the ball is caught. Use the notation x, y, an ordered pair of values separated by a comma, where x and y are expressed numerically in meters, as measured from the point where the softball initially left the bat. Express your answer to three significant figures.

[tex]u_y = v_0 sin \theta[/tex] is the initial vertical velocity, with [tex]v_0[/tex] being the initial speed and [tex]\theta[/tex] the angle of projection

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

t is the time

The ball reaches the ground again when the displacement is zero, s = 0, so we have

[tex]0=v_0 sin \theta t - \frac{1}{2}gt^2[/tex]

So, neglecting the solution for t =0,

[tex]v_0 sin \theta = \frac{1}{2}gt[/tex] (1)

where t = 2.00 s is the time of flight.

At the same time, the third baseman runs backwars at velocity

v = 7.00 m/s

From an initial distance of

L = 18.0 m

So his x-position at time t is

[tex]x_b(t) = L-vt[/tex]

While the x-position of the ball at time t is

[tex]x_s(t)=u_x t = v_0 cos \theta t[/tex]

The player catches the ball when the two positions are the same:

[tex]L-vt=v_0 cos \theta t[/tex] (2)

Dividing eq.(1) by eq.(2) we get

[tex]\frac{tan \theta}{t}=\frac{gt}{2(L-vt)}[/tex]

And so we can use this to find the angle of projection:

[tex]\theta = tan^{-1}(\frac{gt^2}{2(L-vt)})=tan^{-1}(\frac{(9.8)(2.00)^2}{2(18-(7)(2))})=78.5^{\circ}[/tex]

And now we can use eq.(2) to find the initial speed of the ball:

[tex]v_0=\frac{L-vt}{cos \theta t}=\frac{18-(7)(2)}{cos 78.5^{\circ} \cdot 2.00}=10.0 m/s[/tex]

3)

Here we want the velocity of the softball 0.100 s before the ball is caught, so at time

t = 2.00 - 0.100 = 1.90 s

The components of the velocity of the softball are given by:

[tex]v_x = v_0 cos \theta[/tex] which is constant, therefore

[tex]v_x = (10.0)(cos 78.5^{\circ})=2.00 m/s[/tex]

While the y-component of the velocity is given by the equation for accelerated motion:

[tex]v_y = v_0 sin \theta -gt[/tex]

And substituting t = 1.90 s,

[tex]v_y = (10.0)(sin 78.5^{\circ})-(9.80)(1.90)=-8.82 m/s[/tex]

where the negative sign indicates that the direction is downward.

Therefore the vector expression for the velocity of the ball at 0.100 s before it has been caught is

[tex](v_x,v_y)=(2.00,-8.82) m/s[/tex]

4)

The x-position of the ball at time t is given by the equation for uniform motion:

[tex]x=v_0 cos \theta t[/tex]

And substituting t = 1.90 s,

[tex]x=(10.0)(cos 78.5)(1.90)=3.79 m[/tex]

Instead, the vertical position of the ball is given by the equation of uniformly accelerated motion:

[tex]y=v_0 sin \theta t - \frac{1}{2}gt^2[/tex]

And substtuting t = 1.90 s, we find

[tex]y=(10.0)(sin 78.5^{\circ})(1.90)-\frac{1}{2}(9.80)(1.90)^2=0.93 m[/tex]

Of course we have measured the vertical displacement from the head of the player who hit the ball; therefore, we have to add his height, h. So the position of the ball at t = 1.90 s is

[tex](x,y)=(3.79, 0.93+h)m[/tex]

Learn more about projectile motion:

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