Chapter 3: Q45PE (page 124)

In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of $24.77 m$ . What was the initial speed of the shot if he released it at a height of $2.10 m$ and $38.0 °$ threw it at an angle of above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at $45 °$ when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, $38 °$ will give a longer range than $45 °$ in the shot put.)

Short Answer

Expert verified

The initial speed is $15 m / s$

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 $38 °$

$\cos θ = \frac{q}{h} \cos θ = \frac{V_{i x}}{V_{i}} V_{i x} = V_{i} \cos 38$

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

$\bar{V} = \frac{x}{t} V_{i} \cos θ = \frac{24.77}{t} t = \frac{24.77}{V_{i} \cos 38}$

The time taken is used from the above equation.

Putting the values into the equation of motion

role="math" localid="1668677707679" $∆ Y = V_{i y} t + \frac{1}{2} a_{y} t^{2} - 2.1 = (V_{i} \sin θ) t + \frac{1}{2} (- 9.8) t^{2} p u t t h e v a l u e o f t - 2.1 = (V_{i} \sin 38) (\frac{24.77}{V_{i} \cos 38}) - 4.9 {(\frac{24.7}{V_{i} \cos 38})}^{2} - 2.1 = 19.4 - 4.9 {(\frac{24.77}{V_{i} \cos 38})}^{2}$