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Mechanics, EXT2 EQ-Bank 30

In a circus act, an 8 kg cannon ball is projected from the origin into the air with an initial velocity of 26 m s\(^{-1}\) and at an angle of 67.4° to the horizontal. The ball is caught at the top of its trajectory by a performer who is at the position \((A, B)\).

The velocity vector, \(\mathbf{v} (t)\), of the ball at time \(t\) seconds after launch is given by

\(\mathbf{v}(t)=10 e^{-0.8 t} \mathbf{i} +\left[36.5 e^{-0.8 t}-12.5\right] \mathbf{j}\).   (Do NOT Prove this.)

  1. Show that the ball reaches the performer at  \(t=1.339\) (to three decimal places).   (2 marks)

  2. Find the values of \(A\) and \(B\).   (3 marks)

  3. What is the speed of the ball when it reaches the performer?   (1 mark)

  4. What is the magnitude of the force on the ball when it reaches the performer?   (2 marks)

Show Answers Only

a.    \(t=1.339 \ \text{s}\)

b.    \(A=8.217\ \text{m},\ \ B=13.257\ \text{m}\)

c.    \(\text{Speed}\ = 3.426\ \text{m s}^{-1}\)

d.    \(82.951 \ \text{N}\)

Show Worked Solution

a.    \(\text{At top of trajectory:}\)

\(36.5 e^{-0.8 t}-12.5\) \(=0\)
\(36.5 e^{-0.8 t}\) \(=12.5\)
\(e^{-0.8 t}\) \(=\dfrac{12.5}{36.5}\)
\(-0.8 t\) \(=\ln \dfrac{12.5}{36.5}\)
\( t\) \(=\dfrac{\ln\frac{12.5}{36.5}}{-0.8}=1.339 \ \text{s (3 d.p.)}\)

 

b.     \(\text{Horizontal velocity}\ =10 e^{-0.8 t}\)

\(x=\displaystyle \int 10 e^{-0.8 t}\,d t=\dfrac{10}{-0.8} e^{-0.8 t}+c_1=-12.5 e^{-0.8 t}+c_1\) 

\(\text{When} \ \ t=0, \ x=0:\)

\(0=-12.5 e^0+c_1\ \ \Rightarrow\ \ c_1=12.5\)

\(x=-12.5 e^{-0.8 t}+12.5\)
 

\(\text{When } t=1.339, \ x=A:\)

\(A=-12.5 e^{-0.8 \times 1.339}+12.5=8.217\ \text{m (3 d.p.)}\)
 

\(\text {Vertical velocity}\ =36.5 e^{-0.8 t}-12.5\)

\(y=\displaystyle \int\left(36.5 e^{-0.8 t}-12.5\right)\,d t=\dfrac{36.5}{-0.8} e^{-0.8 t}-12.5 t+c_2=-45.625 e^{-0.8 t}-12.5 t+c_2\)

\(\text{When} \ \ t=0, \ y=0:\)

\(0=-45.625+c_2\ \ \Rightarrow\ \ c_2=45.625\)

\(y=-45.625 e^{-0.8 t}-12.5 t+45.625\)
 

\(\text{When}\ \ t=1.339, \ y=B:\)

\(B=-45.625 e^{-0.8 \times 1.339}-12.5 \times 1.339+45.625=13.257\ \text{m (3 d.p.)}\)
 

c.    \(\text{At top of trajectory,}\ \mathbf{j} \text{-component of velocity = 0.}\)

\(\Rightarrow\ \text{Speed at top is the}\ \mathbf{i} \text{-component of velocity at}\ t=1.339:\)

\(\text{Speed}\ =10e^{-0.8 \times 1.339} = 3.426\ \text{m s}^{-1}\ \text{(3 d.p.)}\)
 

d.    \(\text{Using}\ \ F=m \ddot{x}:\)

\(\mathbf{v}(t)=10 e^{-0.8 t}\mathbf{i} +\left(36.5 e^{-0.8 t}-12.5\right) \mathbf{j}\)

\(\mathbf{a} =\dfrac{d v }{d t}=-8 e^{-0.8 t} \mathbf{i} -29.2 e^{-0.8 t} \mathbf{j}\)

\(\text{When} \ \ t=1.339:\)

\(\mathbf{a}=-8 e^{-0.8 \times 1.339}\mathbf{i}-29.2 e^{-0.8 \times 1.339}\mathbf{j}=-2.74 1\,\mathbf{i} -10\, \mathbf{j}\)
 

\(\text{Magnitude of acceleration}\)

\(=\sqrt{(2.741)^2+(10)^2}=10.36885 \ldots\ \text{ms}^{-2}\)
 

\(\therefore \ \text{Magnitude of force}\)

\(=8 \times 10.36885 \ldots =82.951 \ \text{N (3 d.p.)}\)