Question

Problem 4 the bounced ball before the bounce. An outfielder throws a baseball to his catcher in an attempt to throw out a runner at home plate. The ball bounces once before reaching the catcher. Assume the angle at which shown in the figure below, but that the ball's speed after the bounce is one-half of what is was (a) Assume the ball is always thrown with the same initial speed and ignore air l leaves the ground is the same as the angle at which the outfielder threw it as resistance. At what angle 0 should the fielder throw the ball to make it go the same distance D with one bounce (blue path) as a ball thrown upward at 45.0° with no bounce (green path)? (b) Determine the ratio of the time interval for the one-bounce throw to the flight time for the no-bounce throw 45.0° Problem 5 A water hose is used to fill a large cylindrical storage tank of diameter D and height 2D. The hose shoots the water at 45° above the horizontal from the same level as the base of the tank and is a distance 6D away. For what range of lunch speeds vo will the water enter the tank? Ignore air resistance, and express your answer in terms of D and g. 2D Water

Answer 1

4)

With one bounce thrown at an angle $\theta$ , initial speed of ball is $v$. Range of ball is $R_1=\frac{v^2\sin2\theta}{g}$

After one bounce, the ball's velocity is $v/2$ and is making an angle $\theta$ with ground, Range of ball is $R_2=\frac{\left ( \frac{v}{2} \right )^2\sin2\theta}{g}=\frac{v^2\sin2\theta}{4g}$

from the figure, $R_1+R_2=D$  $\Rightarrow D=\frac{5v^2\sin2\theta}{4g}$------(1)

For no bounce, Angle of projection is $45\degree$ , initial speed is $v$.,Range of ball is $D=\frac{v^2\sin2(45\degree)}{g}=\frac{v^2}{g}$ ---(2)

equation (1) and (2) $\frac{5v^2\sin2\theta}{4g}=\frac{v^2}{g}$ $\Rightarrow \theta=\frac{1}{2}\sin^{-1}\left ( \frac{4}{5} \right )=26.6\degree$

b) Time interval for one bounce is $t_{one}=t_1+t_2=\frac{2v\sin\theta}{g}+\frac{2v/2\sin\theta}{g}=\frac{6v\sin\theta}{2g}=0.672\frac{2v}{g}$

Time interval for no bounce is $t_{no}=\frac{2v\sin45\degree}{g}=0.707\frac{2v}{g}$

Ratio of time intervals is $\frac{t_{one}}{t_{no}}=\frac{0.672\frac{2v}{g}}{0.707\frac{2v}{g}}=0.95$

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5)

The water from the hose must reach a horizontal distance between $x=6D$ to $x=7D$ and to a vertical distance of $y=2D$ when it reaches the tank.

Using $s=ut+\frac{1}{2}at^2$

Horizontal distance reached by water in time $t$ is  $x=v_0\sin45\degree t=\frac{v_0t}{\sqrt{2}}$ ---(1)

Vertical distance reached by water in time $t$ is  $y=v_0\sin45\degree t-\frac{1}{2}gt^2$

$y=\frac{v_0 t}{\sqrt{2}}-\frac{1}{2}gt^2$ ----(2)

using (1) and (2) $y=\frac{v_0 (x\sqrt{2}/v_0)}{\sqrt{2}}-\frac{1}{2}g(x\sqrt{2}/v_0)^2$

$y=x-\frac{g}{v_0^2}x^2$

$x-\frac{g}{v_0^2}x^2=2D$ (since $y=2D$)

$\frac{g}{v_0^2}x^2-x+2D=0$

$x^2-\frac{v_0^2}{g}x+\frac{2Dv_0^2}{g}=0$

Solution of above equation is $x=\frac{v_0^2}{2g}\pm\frac{v_0^2}{2g}\sqrt{1-\frac{8gD}{v_0^2}}$

Water reaches tank, when it is travelling in downward direction, so for $6D<x<7D$ , $x=\frac{v_0^2}{2g}&plus;\frac{v_0^2}{2g}\sqrt{1-\frac{8gD}{v_0^2}}$

$6D<\frac{v_0^2}{2g}&plus;\frac{v_0^2}{2g}\sqrt{1-\frac{8gD}{v_0^2}}<7D$

$12gD-v_0^2<{v_0^2}\sqrt{1-\frac{8gD}{v_0^2}}<14gD-v_0^2$

$(12gD-v_0^2)^2<{v_0^4}\left ( {1-\frac{8gD}{v_0^2}} \right )<(14gD-v_0^2)^2$

$(12gD-v_0^2)^2<(v_0^4-8gDv_0^2)<(14gD-v_0^2)^2$

using left identity , $(12gD-v_0^2)^2<(v_0^4-8gDv_0^2)$ $\Rightarrow v_0>\sqrt{9gD}$

using right identity $(v_0^4-8gDv_0^2)<(14gD-v_0^2)^2$   $\Rightarrow v_0<\sqrt{\frac{49}{5}gD}$

Range of speeds lies between  $\sqrt{\frac{49}{5}gD}>v_0>\sqrt{9gD}$

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