Question

A projectile is fired horizontally with a speed of $98{\text{ms}}^{-1}$ from the top of a hill $490\text{m}$ high. Find
(i) The time taken to reach the ground
(ii)The distance of the target from the hill and
(iii)The velocity with which the projectile hits the ground. (take $g=9.8{\text{m/s}}^2$)

• A. (i) $5\text{s}$, (ii) $490\text{m}$, (iii) $110\text{m/s}$
• B. (i) $10\text{s}$, (ii) $980\text{m}$, (iii) $98\sqrt{2}\text{m/s}$
• C. (i) $15\text{s}$, (ii) $1470\text{m}$, (iii) $110\text{m/s}$
• D. (i) $5\text{s}$, (ii) $980\text{m}$, (iii) $98\text{m/s}$

Answer 1

The correct option is B (i) $10\text{s}$, (ii) $980\text{m}$, (iii) $98\sqrt{2}\text{m/s}$



(i) The projectile is fired from the top 0 of a hill with speed $u_x=98{\text{ms}}^{-1}$ and $u_y=0$ along the horizontal as shown. It reaches the target $P$ at vertical depth $OA$, in the coordinate system as shown.
$OA=y=490\text{m}$
From second equation of motion,
$y=u_{y}t+\frac{1}{2}g{t}^2=\frac{1}{2}g{t}^2$
$\Rightarrow 490=\frac{1}{2}\times 9.8t^2$
$\Rightarrow t=\sqrt{100}=10\text{s}$

(ii) Distance of the target from the hill is given by
$AP=x=\text{Horizontal velocity}\times \text{time}=98\times 10=980\text{m}$

(iii) The horizontal and vertical components of velocity $v$ of the projectile at point $P$ are
$v_x=u=98{\text{ms}}^{-1}$
$v_y=u_y+gt=0+9.8\times 10=98{\text{ms}}^{-1}$
$\therefore v=\sqrt{v_x^2+v_y^2}=\sqrt{{98}^2+{98}^2}=98\sqrt{2}{\text{ms}}^{-1}$