Photo AI
A particle is projected from a point on horizontal ground with an initial speed of 25 m/s at an angle $\beta^\circ$ to the horizontal where $\tan \beta = \frac{4}{3}$ - Leaving Cert Applied Maths - Question 3 - 2008
Question 3
A particle is projected from a point on horizontal ground with an initial speed of 25 m/s at an angle $\beta^\circ$ to the horizontal where $\tan \beta = \frac{4}{3}... show full transcript
Worked Solution & Example Answer:A particle is projected from a point on horizontal ground with an initial speed of 25 m/s at an angle $\beta^\circ$ to the horizontal where $\tan \beta = \frac{4}{3}$ - Leaving Cert Applied Maths - Question 3 - 2008
Step 1
Find the initial velocity of the particle in terms of $\mathbf{i}$ and $\mathbf{j}$.
Answer
To determine the initial velocity, we can break down the components based on the angle ( \beta ) and the initial speed of 25 m/s.
The angle can be found using ( \tan \beta = \frac{4}{3} ), leading to ( \beta \approx 53.13^\circ ).
Thus, the components are:
- Horizontal component:
[ v_x = 25 \cos(\beta) \approx 25 \cos(53.13^\circ) = 15 , \text{m/s} ] - Vertical component:
[ v_y = 25 \sin(\beta) \approx 25 \sin(53.13^\circ) = 20 , \text{m/s} ]
Therefore, the initial velocity vector is:
[ \mathbf{v} = 15 \mathbf{i} + 20 \mathbf{j} ]
Step 2
Calculate the time taken to reach the maximum height.
Answer
At maximum height, the vertical velocity becomes zero.
Using the equation ( v = u + at ):
- Initial velocity ( u = 20 , \text{m/s} )
- Final velocity ( v = 0 , \text{m/s} )
- Acceleration ( a = -10 , \text{m/s}^2 ) (due to gravity)
Setting up the equation:
[ 0 = 20 - 10t ]
Solving for ( t ):
[ t = 2 , \text{s} ]
Thus, the time taken to reach maximum height is 2 seconds.
Step 3
Calculate the maximum height of the particle above ground level.
Answer
We can use the equation of motion:
[ s = ut + \frac{1}{2}at^2 ]
Substituting:
- ( u = 20 , \text{m/s} )
- ( t = 2 , \text{s} )
- ( a = -10 , \text{m/s}^2 )
Calculating:
[ s = 20(2) + \frac{1}{2}(-10)(2^2) ]
[ s = 40 - 20 = 20 , \text{m} ]
Therefore, the maximum height above ground level is 20 m.
Step 4
Find the range.
Answer
The range can be determined using the time of flight and horizontal velocity.
The total time taken for the projectile to return to the ground is ( 4 , \text{s} ) (as it goes up and comes down).
Thus, using the range formula:
[ R = v_x \cdot t ]
With ( v_x = 15 , \text{m/s} ) and ( t = 4 , \text{s} ):
[ R = 15 \times 4 = 60 , \text{m} ]
The range is therefore 60 m.
Step 5
Find the speed and direction of the particle after 3 seconds of motion.
Answer
After 3 seconds, we can find the velocity components:
- Horizontal component remains:
( v_x = 15 , \text{m/s} ) - Vertical component after 3 seconds:
[ v_y = u + at = 20 - 10(3) = 20 - 30 = -10 , \text{m/s} ]
Thus, the velocity vector after 3 seconds is:
[ \mathbf{v} = 15 \mathbf{i} - 10 \mathbf{j} ]
To find the speed, we can calculate:
[ \text{speed} = \sqrt{(15)^2 + (-10)^2} = \sqrt{225 + 100} = \sqrt{325} \approx 18.0 , \text{m/s} ]
To find the direction:
[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{-10}{15} ]
[ \theta = \tan^{-1}\left(\frac{-10}{15}\right) \approx -33.69^\circ ]
This indicates the direction is downwards relative to the horizontal.