3. (a) A ball is kicked from a point P on horizontal ground with a speed of 20 m s⁻¹ at 45° to the horizontal. The ball strikes the ground at Q. Find (i) the time it takes the ball to travel from P to Q. (ii) |PQ|, the distance from P to Q. (b) A particle is projected with initial velocity 21𝑖 + 50𝑗 m s⁻¹ from point P on a horizontal plane. A and B are two points on the trajectory (path) of the particle. The particle reaches point A after 3 seconds of motion. The displacement of point B from P is k𝑖 + 80𝑗 metres. Find (i) the velocity of the particle at A in terms of 𝑖 and 𝑗 (ii) the speed and direction of the particle at A (iii) the value of k.

Leaving Cert Applied Maths Projectiles: 3. (a) A ball is kicked from a

3. (a) A ball is kicked from a point P on horizontal ground with a speed of 20 m s⁻¹ at 45° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2012

Question 3

3. (a) A ball is kicked from a point P on horizontal ground with a speed of 20 m s⁻¹ at 45° to the horizontal. The ball strikes the ground at Q. Find (i) the time... show full transcript

Worked Solution & Example Answer:3. (a) A ball is kicked from a point P on horizontal ground with a speed of 20 m s⁻¹ at 45° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2012

Step 1

(i) the time it takes the ball to travel from P to Q

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Answer

To find the time it takes for the ball to reach the ground, we can use the vertical motion equation:

$s = ut + \frac{1}{2}at^2$

Where:

$s$ is the vertical distance (height), which is 0 (ground level)
$u$ is the initial vertical speed, which can be calculated as $20 \sin(45^\circ) = 10\sqrt{2} \
$a$ is the acceleration due to gravity, taken as $-10 m/s^2$ (downwards).

Setting the equation to zero:

$0 = 10\sqrt{2}t - 5t^2$

Factoring out t:

$t(10\sqrt{2} - 5t) = 0$

Thus, the time $t$ when the ball strikes the ground is:

$t = \frac{20\sqrt{2}}{10} = 2\sqrt{2} \approx 2.82 s$

Step 2

(ii) |PQ|, the distance from P to Q

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Answer

To find the distance from P to Q, we can use the horizontal component of the motion:

Since there is no horizontal acceleration:

$x = ut \Rightarrow |PQ| = 20 \cos(45^\circ) \cdot t = 20 \cdot \frac{1}{\sqrt{2}} \cdot 2\sqrt{2}$

This simplifies to:

$|PQ| = 20 \cdot 2 = 40 m$

Step 3

(i) the velocity of the particle at A in terms of 𝑖 and 𝑗

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Answer

The velocity of the particle at A can be calculated using the initial velocity and the acceleration due to gravity:

In the horizontal direction (x-axis):

- In the vertical direction (y-axis): $$ v_y = 50 - 10 \cdot 3 = 20 $$ Thus, the velocity at A is: $$ v = 21𝑖 + 20𝑗 \ m/s $$

Step 4

(ii) the speed and direction of the particle at A

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The speed at point A can be determined by:

$|v| = \sqrt{(21)^2 + (20)^2} = \sqrt{441 + 400} = 29 \ m/s$

The direction (angle $heta$ ) can be found using the tangent function:

$\theta = \tan^{-1}\left(\frac{20}{21}\right) \approx 43^\circ$

Step 5

(iii) the value of k

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Answer

To find the value of k, we look at the kinematic equations. The vertical motion of the particle at point B is:

Substituting t = 3: $$ 80 = 50 - 10(3)^2 \ This gives us 80 after solving $ $$ 80 = 50 - 90 \Rightarrow k = 80 \ + 90 = 168 $$

3. (a) A ball is kicked from a point P on horizontal ground with a speed of 20 m s⁻¹ at 45° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2012

Worked Solution & Example Answer:3. (a) A ball is kicked from a point P on horizontal ground with a speed of 20 m s⁻¹ at 45° to the horizontal - Leaving Cert Applied Maths - Question 3 - 2012

(i) the time it takes the ball to travel from P to Q

(ii) |PQ|, the distance from P to Q

(i) the velocity of the particle at A in terms of 𝑖 and 𝑗

(ii) the speed and direction of the particle at A

(iii) the value of k

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