Question

A projectile is fired with a horizontal velocity of 330m/s of a higher cliffof height 100(H)

(i) How long will it take to strike the level ground at the base of the cliff?

(ii) How far from the foot of the cliff will it strike the ground?

(iii) With what velocity will it strike the ground?

(iv) At what angle will it strike the ground?

Answer 1

Explanations & Calculations

i)

• Apply $\small S= ut +\frac{1}{2}at^2$ for the projectile's vertical motion, downwards $\downarrow$

$\qquad\qquad \begin{aligned} \small 100&=\small \frac{1}{2}gT^2 \end{aligned}$

ii)

• Apply the same formula for its horizontal motion.

$\qquad\qquad \begin{aligned} \small \to s&=\small ut+\frac{1}{2}at^2\\ \small R&=\small 330\,ms^{-1}\times T \end{aligned}$

iii)

• To find the vertical velocity component at the landing, apply $\small v=u+at$ for its vertical motion, again downward.

$\qquad\qquad \begin{aligned} \small \downarrow v_y&=\small 0+gT\to v_y=gT \end{aligned}$

• Then the velocity at the landing is the compound value of both horizontal & vertical velocities.

$\qquad\qquad \begin{aligned} \small V&=\small \sqrt{v_x^2+v_y^2}\\ &=\small \sqrt{330^2+g^2T^2} \end{aligned}$

iv)

• The angle at which it would be landing is

$\qquad\qquad \begin{aligned} \small \tan\theta&=\small \frac{v_y}{v_x}\\ \small \theta&=\small \tan^{-1}\bigg[\frac{gT}{330}\bigg] \end{aligned}$