Question

From the top of a $11\text{m}$ high tower a stone is projected with speed $10\text{m/s}$, at an angle of ${37}^{\circ }$ as shown in the figure. (Take $g=10m/s^2$)


Match the quantities in column-I to their values in column-II

Column-I	Column-II
(i) Speed after $2\text{s}$	p. $8\sqrt{5}$
(ii)Time of flight	q. $12.8$
(iii)Range	r. $\frac{11}{5}$
(iv) The maximum height attained by the stone	s. $\frac{88}{5}$
(v) Speed just before striking the ground.	t. $\sqrt{260}$

• A. (i) - r, (ii) - s, (iii) - p, (iv) - q, (v) - t
• B. (i) - p, (ii) - t, (iii) - s, (iv) - q, (v) - r
• C. (i) - t, (ii) - r, (iii) - s, (iv) - q, (v) - p
• D. (i) - p, (ii) - r, (iii) - s, (iv) - q, (v) - t

Answer 1

The correct option is C (i) - t, (ii) - r, (iii) - s, (iv) - q, (v) - p

a) Initial velocity in horizontal direction $=10\cos {37}^{\circ }=8\text{m/s}$
Initial velocity in vertical direction $=10\sin {37}^{\circ }=6\text{m/s}$
Velocity after 2 seconds
$v=v_x\hat{i}+v_y\hat{j}=8\hat{i}+(u_y+a_{y}t)\hat{j}=8\hat{i}+(6-10\times 2)\hat{j}=8\hat{i}-14\hat{j}$
Speed (magnitude of velocity) $=\sqrt{8^2+{14}^2}=\sqrt{260}\text{m/s}$

b) Using second equation of motion, $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$\Rightarrow -11=6\times t+\frac{1}{2}\times (-10)t^2$
$\Rightarrow 5t^2-6t-11=0\Rightarrow (t+1)(5t-11)=0$
$\Rightarrow t=\frac{11}{5}\text{sec}$ (taking positive value)
c) Range $=u_x\times t=8\times \frac{11}{5}=\frac{88}{5}\text{m}$

d) Maximum height above the level of projection, $h=\frac{u_y^2}{2g}=\frac{6^2}{2\times 10}=1.8\text{m}$

$\therefore$ Maximum height above ground $=11+1.8=12.8\text{m}$

e) Speed just before striking the ground $v=\sqrt{v_x^2+v_y^2}=\sqrt{u_x^2+u_y^2+2gh}=\sqrt{(8{)}^2+(6{)}^2+2\times (-10)\times (-11)}=\sqrt{100+2\times 10\times 11}$
$\Rightarrow v=8\sqrt{5}\text{m/s}$