Question

Suppose the departure angle in Figure 3.26 in the textbook is ?0 = 42.3 ? and the distance is d = 2.90 m . A) Where will the dart and monkey meet if the initial speed of the dart is 12.0 m/s ? B) Where will the dart and monkey meet if the initial speed of the dart is 7.5 m/s ? C) What will happen if the initial speed of the dart is 4.00 m/s? (Response question)

Answer 1

for dart

dart speed components.

horizontal component

$cos\Theta =\frac{V_{ox}}{ V_{0}}\Rightarrow V_{0x}\cdot V_{0}\cdot cos\Theta$

vertical component

$sin\Theta =\frac{V_{oy}}{ V_{0}}\Rightarrow V_{0y}\cdot V_{0}\cdot sin\Theta$

vertical position for the dart

$Y_{d}=Y_{0d}\dotplus V_{oyd}\cdot t\dotplus \frac{1}{2}\cdot a_{y}\cdot t^{2}$

where:

Initial position

$Y_{0d}=0m$
initial vertical velocity

$V_{0dy}=V_{0d}\cdot sin\Theta$
acceleration of gravity

$a_{y}=-g=-9.81\frac{m}{s^{2}}$

Evaluating the equation

$Y_{d}= V_{0d}\cdot sin\Theta\cdot t- \frac{1}{2}\cdot g\cdot t^{2}$

horizontal movement equation

$X_{d}= X_{od}\dotplus V_{0dx}\cdot t$

where:

initial horizontal position

$X_{od}=0m$
initial horizontal velocity

$V_{odx}=V_{odx}\cdot cos\Theta$

finally

$X_{d}= V_{0d}\cdot cos\Theta \cdot t$

for $X_{d}=d$   the time it takes for the dart to reach the monkey is $X_{d}= V_{0d}\cdot cos\Theta \cdot t\Rightarrow t=\frac{d}{V_{0d}\cdot cos\Theta }$

For Monkey

vertical position for the monkey

$Y_{m}=Y_{0m}\dotplus V_{oym}\cdot t\dotplus \frac{1}{2}\cdot a_{y}\cdot t^{2}$

where:

Initial position

$Y_{0d}=d\cdot tan\Theta$
initial vertical velocity

$V_{0ym}=0\frac{m}{s}$
acceleration of gravity

$a_{y}=-g=-9.81\frac{m}{s^{2}}$

Evaluating the equation

$Y_{m}= d\cdot tan\Theta - \frac{1}{2}\cdot g\cdot t^{2}$

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A) Where will the dart and monkey meet if the initial speed of the dart is 12.0 m/s ?

the time are calculated

$t=\frac{d}{V_{0d}\cdot cos\Theta }=\frac{ 2.90m}{12\frac{m}{s}\cdot cos\left ( 42.3^{0} \right ) }=0.33s$

position Monkey and dart

$Y_{d}= 12\frac{m}{s}\cdot sin \left ( 42.3^{0} \right )\cdot 0.33s - \frac{1}{2}\cdot 9.81\frac{m}{s^{2}}\cdot \left ( 0.33s \right )^{2}=2.1m$

$Y_{m}= 2.90m\cdot tan\left ( 42.3^{0} \right ) - \frac{1}{2}\cdot 9.81\frac{m}{s^{2}}\cdot \left ( 0.33s \right )^{2}=2.1m$

They meet at the point $x=2.90m$ and $y=2.1m$

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B) Where will the dart and monkey meet if the initial speed of the dart is 7.5 m/s ?

the time are calculated

$t=\frac{d}{V_{0d}\cdot cos\Theta }=\frac{ 2.90m}{7.5\frac{m}{s}\cdot cos\left ( 42.3^{0} \right ) }=0.52s$

position Monkey and dart

$Y_{d}= 7.5\frac{m}{s}\cdot sin \left ( 42.3^{0} \right )\cdot 0.52s - \frac{1}{2}\cdot 9.81\frac{m}{s^{2}}\cdot \left ( 0.52s \right )^{2}=1.30m$

$Y_{m}= 2.90m\cdot tan\left ( 42.3^{0} \right ) - \frac{1}{2}\cdot 9.81\frac{m}{s^{2}}\cdot \left ( 0.52s \right )^{2}=1.30m$

They meet at the point $x=2.90m$ and $y=1.30m$

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C) What will happen if the initial speed of the dart is 4.00 m/s?

the time are calculated

$t=\frac{d}{V_{0d}\cdot cos\Theta }=\frac{ 2.90m}{4\frac{m}{s}\cdot cos\left ( 42.3^{0} \right ) }=0.98s$

position Monkey and dart

$Y_{d}= 4\frac{m}{s}\cdot sin \left ( 42.3^{0} \right )\cdot 0.98s - \frac{1}{2}\cdot 9.81\frac{m}{s^{2}}\cdot \left ( 0.98s \right )^{2}=-2.1m$

$Y_{m}= 2.90m\cdot tan\left ( 42.3^{0} \right ) - \frac{1}{2}\cdot 9.81\frac{m}{s^{2}}\cdot \left ( 0.98s \right )^{2}=-2.1m$

The results indicate that in this case does not meet,

the vertical position is calculated using both equations to check