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A particle is projected from a point on horizontal ground with an initial speed of 58 ms⁻¹ at an angle β to the horizontal, where tan β = \frac{20}{21} - Leaving Cert Applied Maths - Question 3 - 2011
Question 3
A particle is projected from a point on horizontal ground with an initial speed of 58 ms⁻¹ at an angle β to the horizontal, where tan β = \frac{20}{21}. (i) Find t... show full transcript
Worked Solution & Example Answer:A particle is projected from a point on horizontal ground with an initial speed of 58 ms⁻¹ at an angle β to the horizontal, where tan β = \frac{20}{21} - Leaving Cert Applied Maths - Question 3 - 2011
Step 1
Find the initial velocity of the particle in terms of \( \mathbf{i} \) and \( \mathbf{j} \).
Answer
To find the initial velocity in terms of ( \mathbf{i} ) and ( \mathbf{j} ), we use the components of the initial speed. Given that the speed is 58 ms⁻¹ and ( \tan \beta = \frac{20}{21} ), we can derive the cosine and sine values:
[ \cos \beta = \frac{21}{\sqrt{21^2 + 20^2}} = \frac{21}{29} ] [ \sin \beta = \frac{20}{29} ]
Thus, the initial velocity ( \mathbf{u} ) is:
[ \mathbf{u} = 58 \cos \beta \mathbf{i} + 58 \sin \beta \mathbf{j} = 58 \left(\frac{21}{29}\right) \mathbf{i} + 58 \left(\frac{20}{29}\right) \mathbf{j} = 42 \mathbf{i} + 40 \mathbf{j} ]
Step 2
Calculate the time taken to reach the maximum height.
Answer
At maximum height, the vertical velocity becomes zero. We can use the equation of motion:
[ v = u + at ] Where:
- ( v = 0 ) (final velocity at maximum height)
- ( u = 40 ) (initial vertical velocity)
- ( a = -g = -10 ) (acceleration due to gravity)
Setting up the equation: [ 0 = 40 - 10t ] Solving for ( t ): [ t = 4 ext{ seconds} ]
Step 3
Calculate the maximum height of the particle above ground level.
Answer
To find the maximum height, we can use the equation:
[ s = ut + \frac{1}{2} a t^2 ] Where:
- ( s ) is the maximum height,
- ( u = 40 ),
- ( t = 4 ),
- ( a = -10 ).
Plugging in the values: [ s = 40(4) + \frac{1}{2}(-10)(4^2)\ s = 160 - 80 = 80 ext{ m} ]
Step 4
Find the range.
Answer
The range ( R ) of the projectile can be calculated with:
[ R = u \cdot t + \frac{1}{2} a t^2 ] Using horizontal motion:
- ( u = 42 ) (horizontal initial velocity)
- Total time of flight is ( 2t ) (up and down): ( 2(4) = 8 ) seconds.
Thus: [ R = 42(8) + 0 = 336 ext{ m} ]
Step 5
Find the two times at which the height of the particle is 75 m.
Answer
Using the equation:
[ s = ut + \frac{1}{2} at^2 ] We set ( s = 75 ), ( u = 40 ), and ( a = -10 ):
[ 75 = 40t + \frac{1}{2}(-10)t^2 ] This simplifies to: [ 0 = -5t^2 + 40t - 75 ] Rearranging gives: [ 0 = t^2 - 8t + 15 ] Factoring: [ (t - 3)(t - 5) = 0 ] Thus, the two times are ( t = 3 ) and ( t = 5 ) seconds.