91Ó°ÊÓ

The initial velocity of a shot${}^{{\mathrm{v}}_0=15\mathrm{m}/\mathrm{s}}$

The maximum height reached by the shot${}^{\mathrm{y}=2.16\mathrm{m}}$

The problem deals with the projectile motion of an object. The projectile motion of an object is the motion of an object thrown into the air, subject to the acceleration due to gravity.In this problem, to find the range of the projectile motion at different launching angles kinematic equation of motion an be used.

Formulae:

The newton’s second kinematic equation is,

$∆y=\left(\begin{array}{c}{v}_{0y}\end{array}\right)\left(\begin{array}{c}t\end{array}\right)+\frac{1}{2}a{t}^2$

(i)

The range of the projectile

$$\begin{gathered} R=v_{0x}×t \\ =v_0\cos \mathrm{θ}×t \end{gathered}$$ (ii)

Using equation (i) the maximum height reached by the shot${}^{\left(y\right)}$is,

$$\begin{gathered} ∆y=\left(\begin{array}{c}{v}_{0y}\end{array}\right)\left(\begin{array}{c}t\end{array}\right)-\frac{1}{2}g{t}^2 \\ 0-2.16=15\sin \left(45\right)×t-\left(\frac{1}{2}\right)\left(9.8\right)t^2 \\ -2.16=10.6t-4.9t^2 \end{gathered}$$

Solving this quadratic equation,

$t=2.352s$

With equation (ii) the range of the projectile is

$$\begin{gathered} R=15\cos \left(45\right)×2.352 \\ R=24.95m \end{gathered}$$

Solving for equation (i), the maximum height reached by the shot${}^{\left(y\right)}$is,

$$\begin{gathered} 0-2.16=15\sin \left(42\right)2×t-\left(\frac{1}{2}\right)\left(9.8\right)t^2 \\ -2.16=10.036t-4.9t^2 \end{gathered}$$

Solving this quadratic equation,

localid="1654240192629" $t=2.245s$

The range of the projectile

$$\begin{gathered} \mathrm{R}=15\cos \left(42\right)×2.245 \\ \mathrm{R}=25.02\mathrm{m} \end{gathered}$$

Therefore, using the angle of launch, and the velocity of launch, it is possible to find the range of the shot put. It can be used to optimize the throw angle to achieve the maximum range.