Question

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of $\mathbf{24}\mathbf{.}\mathbf{77}\mathit{m}$. What was the initial speed of the shot if he released it at a height of$\mathbf{2}\mathbf{.}\mathbf{10}\mathit{m}$ and $\mathbf{38}\mathbf{.}\mathbf{0}\mathbf{°}$threw it at an angle of above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at $\mathbf{45}\mathbf{°}$ when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus,$\mathbf{38}\mathbf{°}$ will give a longer range than$\mathbf{45}\mathbf{°}$ in the shot put.)

Short answer

The initial speed is$15m/s$

Step-by-step solution

Stating given data

The shotput is released at the height of$24.77$ meters. The angle of the release of the ball is $38°$

$$\begin{gathered} \cos \theta =\frac{q}{h} \\ \cos \theta =\frac{V_{ix}}{V_i} \\ {V}_{ix}=V_i\cos 38 \end{gathered}$$

The initial velocity in the y direction will be

Calculating angle at which ball should be thrown

The velocity in the x frame will be constant in all time. Considering the initial velocity

$$\begin{gathered} \overline{V}=\frac{x}{t} \\ {V}_i\cos \theta =\frac{24.77}{t} \\ t=\frac{24.77}{V_i\cos 38} \end{gathered}$$

The time taken is used from the above equation.

Putting the values into the equation of motion

role="math" localid="1668677707679" $$\begin{gathered} ∆Y=V_{iy}t+\frac{1}{2}{a}_y{t}^2 \\ -2.1=\left(V_i\sin \theta \right)t+\frac{1}{2}(-9.8)t^2 \\ putthevalueoft \\ -2.1=\left(V_i\sin 38\right)\left(\frac{24.77}{V_i\cos 38}\right)-4.9{\left(\frac{24.7}{V_i\cos 38}\right)}^2 \\ -2.1=19.4-4.9{\left(\frac{24.77}{V_i\cos 38}\right)}^2 \end{gathered}$$

$$\begin{gathered} -2.1=19.4-4.9\left(\frac{988}{{V_i}^2}\right) \\ -2.1=19.4-\frac{4840}{{V_i}^2} \\ {V_i}^2=\frac{-4840}{-21.5}=225 \\ {V}_i=15m/s \end{gathered}$$

The initial speed is 15m/s