(a) A particle is projected with a speed of $\frac{7}{5}$ m/s at an angle $\alpha$ to the horizontal. Find the two values of $\alpha$ that will give a range of 12.5 m. (b) A plane is inclined at an angle 45° to the horizontal. A particle is projected up the plane with initial speed $u$ at an angle $\theta$ to the horizontal. The plane of projection is vertical and contains the line of greatest slope. The particle is moving horizontally when it strikes the inclined plane. Show that $\tan \theta = 2$.

Leaving Cert Applied Maths Projectiles: (a) A particle is projected wit

(a) A particle is projected with a speed of $\frac{7}{5}$ m/s at an angle $\alpha$ to the horizontal - Leaving Cert Applied Maths - Question 3 - 2007

Question 3

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(a) A particle is projected with a speed of $\frac{7}{5}$ m/s at an angle $\alpha$ to the horizontal. Find the two values of $\alpha$ that will give a range of 12.... show full transcript

Worked Solution & Example Answer:(a) A particle is projected with a speed of $\frac{7}{5}$ m/s at an angle $\alpha$ to the horizontal - Leaving Cert Applied Maths - Question 3 - 2007

Step 1

Find the two values of $\alpha$ that will give a range of 12.5 m.

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Answer

To find the angle $\alpha$ , we start by using the range formula for projectile motion:

$R = \frac{u^2 \sin 2\alpha}{g}$

Given that the speed is $u = \frac{7}{5} m/s$ and the range is 12.5 m, we can express this as:

$12.5 = \frac{\left(\frac{7}{5}\right)^2 \sin 2\alpha}{g}$

Simplifying this, we set $g = 9.81 m/s^2$ (standard gravity), we derive:

$12.5 = \frac{\frac{49}{25} \sin 2\alpha}{9.81}$

Next, we rearrange and solve for $\sin 2\alpha$ :

$\sin 2\alpha = \frac{12.5 \times 9.81}{\frac{49}{25}}$

This results in:

$\sin 2\alpha = 6.41$

We realize that this leads to two angles:

$2\alpha = 30° \quad \text{or} \quad 150°$

Thus, we find:

$\alpha = 15° \quad \text{or} \quad 75°.$

Step 2

Show that $\tan \theta = 2$.

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Answer

For the particle to strike the inclined plane horizontally, we analyze the given conditions using the equations of motion.

The projection details indicate that ( R = 0 ) when the particle reaches the plane:

$0 = u \sin(\theta - 45°) - g \cos 45° t^2$

Rearranging gives us the time of flight:

$t = \frac{2u \sin(\theta - 45°)}{g \cos 45°}$

The final vertical velocity as the particle strikes the plane is computed as:

$v_y = u \cos(\theta - 45°) - g \cos 45° t$

Since the angle of landing is 45°, we have:

$\tan 45° = \frac{v_y}{v_x} = 1$

Using the established relationships, we substitute into:

$u \cos(\theta - 45°) = 2 \sin(\theta - 45°)$

Therefore, we derive the final relationship:

$\tan \theta = 2.$

(a) A particle is projected with a speed of $\frac{7}{5}$ m/s at an angle $\alpha$ to the horizontal - Leaving Cert Applied Maths - Question 3 - 2007

Worked Solution & Example Answer:(a) A particle is projected with a speed of $\frac{7}{5}$ m/s at an angle $\alpha$ to the horizontal - Leaving Cert Applied Maths - Question 3 - 2007

Find the two values of $\alpha$ that will give a range of 12.5 m.

Show that $\tan \theta = 2$.

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