Everyday I'm jugglin'

Let the initial speed be $v$ . Let the horizontal component be $v_x$ and the vertical component be $v_y$ . We know that $v^2 = v_x^2 + v_y^2,$ so we need to obtain $v_x$ and $v_y$ , so that we can calculate $v$ .

Vertical Motion : We have that $v_y^2 = 2gh.$ This is simply because the $y$ -velocity at the peak is $0$ . We will need to find the time of flight for future reference. To do this, note that $h = \dfrac {1}{2}gt^2,$ where $t$ is the time to reach the peak of the trajectory. We obtain $t = \sqrt {\dfrac {2h}{g}}.$ By symmetry, the time of flight is $2t = 2 \cdot \sqrt {\dfrac {2h}{g}}$ .

Horizontal Motion : The projectile travels horizontally a distance $d = 0.50 \, \text{m}$ in a time $2t$ . Thus, the horizontal velocity is $v_x = \dfrac {d}{2t} = \dfrac {d}{2 \cdot \sqrt {\dfrac {2h}{g}}}.$

Putting it together : $\begin{aligned} v &= \sqrt {v_x^2 + v_y^2} \\&= \sqrt {\left( \dfrac {d}{2 \cdot \sqrt {\dfrac {2h}{g}}} \right)^2 + 2gh} \\&= \sqrt {\dfrac {gd^2}{8h} + 2gh}. \end{aligned}$ Using $d = 0.50 \, \text{m}$ , $g = 9.8 \, \text{m/s}^2$ , and $h = 0.784 \, \text{m}$ , we have $v = \boxed {\text {3.97 m/s}}.$

Alternatively, we could have just numerically calculated $v_x$ and $v_y$ separately and then used $v = \sqrt {v_x^2 + v_y^2}$ , but the above also works.

In the y component, we know that vf^2-vi^2=2ad and at the top of the parabola vf=0

We get -vi^2=(2)(-9.8)(0.784)

vi=3.92

in the x component, since the time for 2 tosses is 0.4s, then the amount for 4 tosses, which is the period, must be 0.8s.

vx=0.5/0.8

=0.625

v^2 = vi^2 + vx^2

v^2 =3.92^2 + 0.625^2

Therefore, v=3.97

BY USING THE FOLLOWING EQUATION WE CAN FIND THE INITIAL VELOCITY S=UT+1/2AT2 0.784=0.4U-1/2 9.8 0.4*0.4 0.784=0.4U-0.784 U=1.568/0.4 WE GET U AS 3.97

Since the time between two throws is 0.4, the time of flight is 4*0.4=1.6s. Let $\theta$ be the angle at which the ball is thrown. Then, we have the following equations-

$u{\cos(\theta)}t=d$

${1 \over 2}m(u\sin(\theta))^2=mgh$

Solving, for $u$ , we get $u=3.9324$ .