91Ó°ÊÓ

It is given that,

The mass of the shot is M=7.26 kg

The distance travelled by the shot x=1.65 m

The force applied on the shot F=380 N.

The initial velocity of the shot$v_0=2.5m/s$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagram and also kinematic equation, the final velocity of the shot at different anglescan be found.

Formulae:

Newton’s second law is,

$F_{net}=\underset{}{∑}Ma$ (i)

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

Newton’s third kinematic equation is,

$v_f^2=v_0^2+2a\left(∆x\right)$ (ii)

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

FBD for the system is,

From FBD, write for the net force acting in x-direction as,

$$\begin{gathered} F_{netx}=F-Mg\sin \mathrm{θ} \\ {F}_{netx}=380-(7.260)(9.8)\sin 30° \\ {F}_{netx}=344.4N \end{gathered}$$

Using equation (i),

$$\begin{gathered} a_x=\frac{F_{netx}}{M} \\ {a}_x=\frac{344.4}{7.26} \\ {a}_x=47.44m/s^2 \end{gathered}$$

Now, with equation (ii), final velocity will be,

$v_f=\sqrt{{v_0}^2+2a\left(∆x\right)}$

So,the speed of the shot at the end of the acceleration phase if the angle between the path and the horizontal 30° is

$$\begin{gathered} v_f=\sqrt{2.5^2+2\left(47.44\right)\left(1.65\right)} \\ {v}_f=12.76m/s \end{gathered}$$

So, the speed of the shot at the end of the acceleration phase if the angle between the path and the horizontal 30° is $v_f=12.76m/s$

According to Newton’s second law,

At $\mathrm{θ}={42}^o$

$$\begin{gathered} F_{netx}=F-Mg\sin \mathrm{θ} \\ {F}_{netx}=380-\left(7.260\right)\left(9.8\right)\sin 42° \\ {F}_{netx}=332.4N \end{gathered}$$

But,

$a_x=\frac{F_{netx}}{M}$

So,

$$\begin{gathered} a_x=\frac{332.4}{7.26} \\ {a}_x=45.78m/s^2 \end{gathered}$$

So, the speed of the shot at the end of the acceleration phase if the angle between the path and the horizontal is 42° is

$$\begin{gathered} v_f=\sqrt{2.5^2+2\left(45.78\right)\left(1.65\right)} \\ {v}_f=12.54m/s \end{gathered}$$

So, the speed of the shot at the end of the acceleration phase if the angle between the path and the horizontal is 42° is 12.54 m/s.

The percentage decrease in the launch speed if the athlete increases the angle from30°to 42° is,

$\frac{12.76-12.54}{12.76}×100=1.724\%$

The percentage decrease in the launch speed if the athlete increases the angle from30°to 42° is 1.724%.