A projectile is fired with initial velocity 14 i + 10 j m/s from the top of a vertical cliff of height 40 m - Leaving Cert Applied Maths - Question 3 - 2007

The maximum height (H) can be calculated using the formula:

$H = (u_y t) + \frac{1}{2}at^2$

Substituting the known values:

$H = (10)(1) + \frac{1}{2}(-10)(1^2)$

This simplifies to:

$H = 10 - 5 = 5 ext{ m}$

So, the total maximum height above ground level will be:

$40 + 5 = 45 ext{ m}$

Using the motion equation again, we can calculate the distance (s) the projectile falls from the max height to the ground:

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

Here, the initial velocity (u) at maximum height is 0 m/s:

$40 = 0t + \frac{1}{2}(-10)t^2$

Then, we have:

$40 = -5t^2$

Which leads to:

$t^2 = 8$

So, $t = \sqrt{8} = 2.83$ seconds, but including the time to reach the max height (1s), total time = 1 + 2.83 = 3.83 seconds.

To calculate the range, we first find the total time of flight which is 1 second to max height and 3 seconds to ground:

$t_{total} = 1 + 3 = 4s$

Then we can use the horizontal component (u_x = 14 m/s) to find the range (R):

$R = u_x t_{total}$

Thus:

$R = 14 \times 4 = 56 ext{ m}$

To find the speed as it strikes the ground, we use the equation:

$v^2 = u^2 + 2as$

Where:

• Final vertical speed (v) we are calculating
• Initial vertical speed (u) = 0 m/s
• Distance (s) = 40 m
• Acceleration (a) = 10 m/s² (gravity)

Thus,

$v^2 = 0 + 2(10)(40) = 800$

So, $v = \sqrt{800} = 28.28$ m/s for the vertical. The overall speed will be:

$\sqrt{14^2 + (28.28)^2} \approx 31.11 ext{ m/s}$