Question

Suppose that a shot putter can put a shot at the world-class speed${}^{\mathbf{15}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$and at a height of${}^{\mathbf{2}\mathbf{.}\mathbf{160}\mathbf{}\mathbf{m}}$. What horizontal distance would the shot travel if the launch angle${}^{{\mathit{\theta }}_{\mathbf{o}}}$is (a)${}^{\mathbf{45}\mathbf{.}\mathbf{00}\mathbf{°}}$and(b)${}^{\mathbf{42}\mathbf{.}\mathbf{00}\mathbf{°}}$? The answers indicate that the angle of${}^{\mathbf{45}\mathbf{°}}$, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.

Short answer

(a) The horizontal distance traveled by the shot with launch angle${}^{45°\mathrm{is}24.95\mathrm{m}.}$

(b) The horizontal distance traveled by the shot with launch angle${}^{42°\mathrm{is}25.02\mathrm{m}.}$

Step-by-step solution

Given information

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

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

Determining the concept of projectile motion

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

$$\begin{gathered} R=v_{0x}\times t \\ =v_0\cos \theta \times t \end{gathered}$$ (ii)

(a) Determining the horizontal distance travelled by the shot by the shot with launch angle45°

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

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

Solving this quadratic equation,

$t=2.352s$

With equation (ii) the range of the projectile is

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

(b) Determining the horizontal distance travelled by the shot with launch angle42°

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

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

Solving this quadratic equation,

localid="1654240192629" $t=2.245s$

The range of the projectile

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

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.