In a certain track and field event, the shotput has a mass of 7.30 kg and is released with a speed of $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}\mathbf{at}\mathbf{}\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{°}$above the horizontal over a competitor’s straight left leg. What are the initial horizontal and vertical components of the momentum of this shotput?

The mass is, m =7.30kg

The speed is, v =15.0 m/s

And angle is,$\theta =40°$

A characteristic of a moving body that it possesses due to its mass and motion that is broadly equivalent to the product of its mass and velocity: a characteristic of a moving body that defines how long it will take for it to come to rest when subjected to a constant force.

The horizontal component of the initial moment is expressed as,

$\overrightarrow{p_x}=m{\overrightarrow{v}}_x=mv\cos \theta$ …(1)

Substitute all the value in the equation (1)

$$\begin{gathered} p_x=mv\cos \theta \\ =7.30\mathrm{kg}\times 15\mathrm{m}/\mathrm{s}\times \cos 40° \\ =83.88\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The vertical component of the initial moment is expressed as,

$\overrightarrow{p_x}=m{\overrightarrow{v}}_y=mv\sin \theta$ …(2)

Substitute all the value in the equation (2)

$$\begin{gathered} {\mathrm{p}}_{\mathrm{x}}=\mathrm{mv}\mathrm{sin\theta } \\ =7.30\mathrm{kg}\times 15\mathrm{m}/\mathrm{s}\times \sin 40° \\ =70.4\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence the horizontal and vertical component of the initial moment are 83.88 kg.m/s and 70.40kg.m/s