A particle is projected with initial velocity 72 i + 30 j ms⁻¹ from the top of a straight vertical cliff of height 35 m - Leaving Cert Applied Maths - Question 3 - 2010

To find the maximum height above ground level, we use the equation:

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

Where:

• $u = 30$ m/s
• $t = 3$ s
• $a = -10$ m/s²

Now substituting the values:

s = 90 - 45 = 45 ext{ m}$$ The maximum height of the particle above the ground level is **45 m**. The total height from the ground is **80 m** (including the cliff height of 35 m).

To calculate the total time of flight, we must analyze both the upward and downward motions. The upward motion takes 3 seconds, and we need to find the time it takes to fall from the maximum height to the ground. Using the equation:

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

Where:

• $s = -35$ m (since the particle falls 35 m)
• $u = 30 - 10(3) = 0$ m/s (initial velocity at the peak)
• $a = -10$ m/s²

Setting up the equation:

-35 = -5t^2 \\ t^2 = 7 \\ t = \sqrt{7} ext{ s}$$ The total time of flight, adding both segments, is approximately **7 seconds**.

To find the distance |OP|, we can use the horizontal motion:

$|OP| = ut + ut_{total}$

Where:

• $u = 72$ m/s (horizontal velocity)
• $t_{total} = 7$ s (total time of flight)

Calculating the horizontal distance:

$|OP| = 72(7) = 504 ext{ m}$

Thus, |OP|, the distance from O to P, is 504 m.

To find the speed of the particle just before it strikes the ground, we use:

$v_y = u + at$

Where:

• $v_y$ = final vertical velocity
• $u = 30$ m/s
• $t = 7$ s
• $a = -10$ m/s²

Substituting values gives:

$v_y = 30 - 10(7) = 30 - 70 = -40 ext{ m/s}$

For the total speed, we can calculate:

v = \sqrt{5184 + 1600} \\ v = \sqrt{6784} \approx 82.4 ext{ m/s}$$ Thus, the speed of the particle as it strikes the ground is approximately **82.4 m/s**.