Question

As shown in the figure, a projectile is fired with a horizontal velocity of $330\text{m/s}$ from the top of a cliff $80\text{m}$ high. (a) How long will it take for the projectile to strike the level ground at the base of the cliff? (b) How far from the foot of the cliff will it strike? (c) With what velocity will it strike?
(Take $g=10m/s^2$)

• A. (a) $16\text{s}$ (b) $5280\text{m}$ (c) $366.7\text{m/s}$
• B. (a) $4\text{s}$ (b) $1320\text{m}$ (c) $330\text{m/s}$
• C. (a) $4\text{s}$ (b) $1320\text{m}$ (c) $332.4\text{m/s}$
• D. (a) $8\text{s}$ (b) $2640\text{m}$ (c) $332.4\text{m/s}$

Answer 1

The correct option is C (a) $4\text{s}$ (b) $1320\text{m}$ (c) $332.4\text{m/s}$

Let $H$ be the height of the cliff from the ground.
a) Time taken by the projectile to reach ground $T=\sqrt{\frac{2H}{g}}$
$\Rightarrow T=\sqrt{\frac{2\times 80}{10}}=4\text{s}$
b) Horizontal distance covered by the projectile
$x=u_x\times T$
$x=330\times 4=1320\text{m}$
c) Vertical velocity of the projectile when it reach the ground $v_y=u_y+gt$
$v_y=0+10\times 4=40\text{m/s}$
Velocity in the horizontal direction will remain same as there is no acceleration in this direction.
$\Rightarrow v_x=u_x=330\text{m/s}$
Velocity with which the projectile will strike the ground =
$v=\sqrt{v_x^2+v_y^2}$
$=\sqrt{{330}^2+{40}^2}$
$=332.4{\text{ms}}^{-1}$