Hello, Cathaloc2!

Quote:

(A) A projectile is launched horizontally out to sea

with an initial velocity of 98m/s from the top of a vertical cliff 78.4 m high.

(B) The angle of the launch is now raised to 30° above the horizontal.

You need the projectile formulas:

. . $\displaystyle x \;=\; (v_o\cos\theta)t$

. . $\displaystyle y \;=\; h_o + (v_o\sin\theta)t - 4.9t^2$

where: .$\displaystyle \begin{array}{ccc}h_o &=& \text{initial height} \\ v_o &=& \text{initial speed} \\ \theta &=& \text{angle of elevation} \end{array}$

We are given: .$\displaystyle h_o = 78.4,\quad v_o = 98,\quad\theta = 30^o$

The equations are:

. . $\displaystyle x \:=\:(98\cos30^o)t \quad\Rightarrow\quad {\color{blue} x(t) \:=\:49\sqrt{3}\,t}$

. . $\displaystyle y \:=\:78.4 + (98\sin30^o)t - 4.9t^2 \quad \Rightarrow\quad{\color{blue}y(t) \:=\:78.4 + 49t - 4.9t^2}$

Quote:

(i) What is the time taken to reach the sea?

When is $\displaystyle y = 0$ ?

We have: .$\displaystyle 78.4 + 49t - 4.9t^2 \:=\:0\quad\Rightarrow\quad 4.9t^2 - 49t - 78.4 \:=\:0$

Quadratic Formula: .$\displaystyle t \;=\;\frac{49 \pm\sqrt{49^2 - 4(4.9)(-78.4)}}{2(4.9)} \;=\;\frac{49 \pm\sqrt{3937.64}}{9.8}$

The positive root is: .$\displaystyle t \:\approx\:\boxed{11.4 \text{ seconds}}$

Quote:

(ii) How far from the base of the cliff will it hit the sea?

$\displaystyle x(11.4) \;=\;49\sqrt{3}(11.4) \;\approx\;\boxed{967.8\text{ m}}$

Quote:

(iii) How long will it take to reach its greatest height above the point of launch?

The height function, $\displaystyle y \:=\:-4.9t^2 + 49t + 78.4$, is a down-opening parabola.

. . Its maximum is at its *vertex*: .$\displaystyle v \,=\,\frac{-b}{2a}$

We have: .$\displaystyle t \:=\:\frac{-49}{2(-4.9)} \:=\:\boxed{5\text{ seconds}}$

Quote:

(iv) What is the greatest height above the point of launch?

$\displaystyle y(5) \;=\;78.4 + 49(5) - 4.9(5^2) \;=\;\boxed{200.9\text{ m}} $