91Ó°ÊÓ

• The mass of the cannonball M=85 kg.
• The angle with which the cannonball is shot $\mathrm{θ}=53°$
• The range of the motion of the cannonball x=69 m.
• The displacement of the ball in the barrel is 52 m

The kinematic equations relate to the initial velocity, final velocity, acceleration, displacement, and time. The force acting on any object is equal to the product of mass and acceleration of the object; this is referred to as Newton's second law of motion.

Using Newton's kinematic equations of motion, we can find the acceleration of the ball. The magnitude of the force propelling the cannonball in the barrel can be found by substituting the acceleration in the formula for Newton's second law of motion.

Formulae:

According to Newton's second kinematic equation,

$∆x=\left(V_i\right)\left(t\right)+\frac{1}{2}a{t}^2$

According to Newton's first kinematic equation,

$V_f=V_i+at$

According to Newton's third kinematic equation,

$V_f^2=V_i^2+2a\left(∆x\right)$

Newton's second law is,

$F_{net}+=\underset{}{∑}Ma$

Here, $∆x$is displacement,$V_f$is the final velocity, $V_i$is initial velocity, t is time,a is acceleration,$F_{net}$is the net force, and Mis mass of the object.

According to Newton's second kinematic equation,

$∆x=\left(V_i\right)\left(t\right)+\frac{1}{2}a{t}^2$

The range of the projectile motion of the ball is,

$$\begin{gathered} x=V_{ix}t \\ =V_i\cos \mathrm{θ}×t \\ =V_i\cos \left(53°\right)×t \end{gathered}$$

(1)

According to Newton's first kinematic equation,

$V_f=V_i+at$

When the ball reaches at the maximum height, its velocity becomes zero. For the first half of the projectile motion, we can write as,

$$\begin{gathered} 0=V_i\sin \mathrm{θ}-gt \\ {V}_i\sin \mathrm{θ}=gt \\ t=\frac{V_i\sin \mathrm{θ}}{g} \end{gathered}$$

Therefore, the total time of flight is,

$t=\frac{2V_i\sin \mathrm{θ}}{g}$

Inserting this value of t in equation 1 we got,

$$\begin{gathered} x=V_i\cos \left(53°\right)×\frac{2V_i\sin \left(53°\right)}{g} \\ =\frac{V_i^2\sin \left(2\mathrm{θ}\right)}{g} \end{gathered}$$

The initial velocity of the ball during projectile motion is,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{i}}=\sqrt{\frac{\mathrm{gx}}{\sin \left(2\mathrm{θ}\right)}} \\ =\sqrt{\frac{9.8×69}{\sin \left(106°\right)}} \\ =26.52\mathrm{m}/\mathrm{s} \end{gathered}$$

The horizontal and vertical components of the initial velocity of the ball are,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ix}}={\mathrm{V}}_{\mathrm{i}}\mathrm{³¦´Ç²õθ} \\ =26.52\cos \left(53°\right) \\ =15.96\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{iy}}={\mathrm{V}}_{\mathrm{i}}\mathrm{²õ¾\pm ²Ôθ} \\ =26.52\sin \left(26.52°\right) \\ =21.18\mathrm{m}/\mathrm{s} \end{gathered}$$

According to Newton's third kinematic equation,

So, the horizontal and vertical components of acceleration of the ball are,

$$\begin{gathered} {\mathrm{a}}_{\mathrm{x}}=\frac{{\mathrm{V}}_{\mathrm{ix}}^2}{2\mathrm{x}} \\ =\frac{{\left(15.96\mathrm{m}/\mathrm{s}\right)}^2}{2\left(5.2\mathrm{m}\right)} \\ =40.7\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

$$\begin{gathered} {\mathrm{a}}_{\mathrm{y}}=\frac{{\mathrm{V}}_{\mathrm{iy}}^2}{2\mathrm{y}} \\ =\frac{{\left(21.18\mathrm{m}/\mathrm{s}\right)}^2}{2\left(5.2\mathrm{m}\right)} \\ =54.0\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the x and y components of the acceleration are $40.17\mathrm{m}/{\mathrm{s}}^2$and $54.0\mathrm{m}/{\mathrm{s}}^2$, respectively.

Newton's second law is,

$F_{net}=\underset{}{∑}Ma$

So, the horizontal and vertical components of force acting on the ball are,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{x}}={\mathrm{Ma}}_{\mathrm{x}} \\ =85\mathrm{kg}×40.7\mathrm{m}/{\mathrm{s}}^2 \\ =3460\mathrm{N} \end{gathered}$$

$$\begin{gathered} F_y=M{a}_y+Mg \\ =\left(85+54\right)+85×9.8 \\ =5424\mathrm{N} \end{gathered}$$

The magnitude of the force propelling the cannonball in the barrel is,

$$\begin{gathered} \mathrm{F}=\sqrt{{\mathrm{F}}_{\mathrm{x}}^2+{\mathrm{F}}_{\mathrm{y}}^2} \\ =\sqrt{{3460}^2+{5424}^2} \\ =6434\mathrm{N} \\ ≈6.4×{10}^3\mathrm{N} \end{gathered}$$

Therefore, the magnitude of the force propelling the cannonball in the barrel is, $6.4×{10}^3\mathrm{N}$.