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A particle is projected with speed \( \sqrt{\frac{9gh}{2}} \) from a point \( P \) on the top of a cliff of height \( h \) - Leaving Cert Applied Maths - Question 3 - 2017
Question 3
A particle is projected with speed \( \sqrt{\frac{9gh}{2}} \) from a point \( P \) on the top of a cliff of height \( h \). It strikes the ground a horizontal dist... show full transcript
Worked Solution & Example Answer:A particle is projected with speed \( \sqrt{\frac{9gh}{2}} \) from a point \( P \) on the top of a cliff of height \( h \) - Leaving Cert Applied Maths - Question 3 - 2017
Step 1
Find the two possible angles of projection.
Answer
The horizontal distance ( R ) covered by the particle is given as ( R = 3h ).
From the projectile motion equations, we have:
[ R = u \cos \alpha \cdot t ]
where ( \alpha ) is the angle of projection and ( t ) is the total time of flight.
The height equation is given by:
[ h = u \sin \alpha imes t - \frac{1}{2} g t^2 ]
Solving these two equations, we can substitute for ( t ) from the first equation into the second, yielding two possible angles of projection:
[ \tan \alpha = 3 \Rightarrow \alpha = 71.6^{\circ} ]
and
[ \tan \alpha = 0 \Rightarrow \alpha = 0^{\circ} ]
Thus, the two possible angles are ( 0^{\circ} ) and ( 71.6^{\circ} ).
Step 2
For each angle of projection find, in terms of h, the time it takes the particle to reach P.
Answer
Using the first angle ( \alpha = 71.6^{\circ} ):
From the time of flight equation, we can find ( t ):
[ R = u \cos \alpha imes t = 3h ]
We substitute [ u = \sqrt{\frac{9gh}{2}} ]
[ 3h = \sqrt{\frac{9gh}{2}} \cos(71.6^{\circ}) \cdot t ]
Solving for ( t ):
[ t = \frac{3h}{\sqrt{\frac{9gh}{2}} \cdot \cos(71.6^{\circ})} ]
Finally, we can express this in terms of ( h ).
For the angle ( \alpha = 0^{\circ} ):
The total time will just be the vertical drop to height ( h ):
[ t = \sqrt{\frac{2h}{g}} ]
Step 3
Show that the range on the inclined plane is \( \frac{2u^{2} \cos^{2} \theta \sin \theta }{g \cos^{2} \theta} \).
Answer
For a projectile on an inclined plane, the range ( R ) can be derived using the projectile equations. We start with the following:
[ R = \frac{u^2 \sin(2\theta)}{g \cos^2 \theta} ]
Using the identity for ( \sin(2\theta) = 2 \sin \theta \cos \theta ), we substitute this into our formula to show that:
[ R = \frac{2u^2 \sin \theta \cos^2 \theta}{g \cos^2 \theta} ]
Proving the desired relationship.
Step 4
If the particle strikes the inclined plane at right angles show that the range is \( \frac{u^{2}}{g \sqrt{3}} \).
Answer
When a particle strikes the inclined plane at right angles, we can apply conservation of momentum. The velocity components can be calculated at impact. Since the angle is vertical at impact, we have:
[ R = \frac{u^2}{g \sin \theta} ]
Substituting for ( \sin \theta ) gives:
[ R = \frac{u^2}{g \cdot (\frac{1}{\sqrt{3}})} = \frac{u^{2}}{g \sqrt{3}} ]
This shows that the range is as required.