91Ó°ÊÓ

It is given that,

The mass of the shot is$M=7.26\mathrm{kg}$

The distance travelled by the shot$x=1.65\mathrm{m}$

The angle at which the shot is launched,$\mathrm{θ}=34.10°$

The initial velocity of the shot $v_0=2.5\mathrm{m}/\mathrm{s}$

Shot leaves the hand at a height $y_0=2.110\mathrm{m}$

This problem deals with projectile path of an object. In projectile motion the path traced by the object is called trajectory. Also, this problem involves Newton’s Second law of motion. Using the Newton’s second law and kinematic equation of motion, the magnitude of the athlete’s average force on the shot during the acceleration phase can be calculated.

Formulae:

The equation for projectile path is,

$y=y_0+x\tan {\mathrm{θ}}_0-\frac{g{x}^2}{2{\left(V_0\cos {\mathrm{θ}}_0\right)}^2}$ (i)

The Newton’s second law is,

${\mathrm{F}}_{\mathrm{net}}=\underset{}{∑}\mathrm{Ma}$ (ii)

where, $F_{net}$ is the net force, Mis mass and a is an acceleration.

The velocity in Newton’s third kinematic equation is given by,

$v_f^2=v_0^2+2a\left(∆x\right)$ (iii)

where,$v_f$ is the final velocity,$v_0$ is the initial velocity, a is an acceleration and$∆x$ is displacement.

With the given values the equation (i) can be written as,

$$\begin{gathered} 0-2.11=15.9\tan \left(34.10\right)-\frac{\left(9.8\right){\left(15.9\right)}^2}{2{\left(v_0\cos \left(34.10\right)\right)}^2} \\ -2.11=10.769-\frac{2477.54}{2v_0^2\left(0.686\right)} \\ {v}_0^2=140.3 \\ {v}_0=11.85 \end{gathered}$$

Using equation (iii) the acceleration can eb written as,

$$\begin{gathered} a=\frac{v_f^2-v_0^2}{2\left(∆x\right)} \\ a=\left[{\left(11.85\right)}^2-{\left(2.5\right)}^2\right]/\left[2\left(1.65\right)\right] \\ a=40.63\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of the athlete’s average force on the shot during the acceleration phase is,

$$\begin{gathered} F=Ma+Mg\sin \mathrm{θ} \\ F=7.26\left(40.63+9.8×\sin 34.10°\right) \\ F=334.8\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the athlete’s average force on the shot during the acceleration phase is 334.8 N .