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 ${37}^{\circ }$ with the horizontal as shown in the figure. Find
a) Time of flight.
b) Horizontal range.
c) Speed just before striking the ground.
(Take $g=10m/s^2$)

• A. (a) $\frac{11}{5}\text{s}$, (b) $\frac{88}{5}\text{m}$, (c) $8\sqrt{5}\text{m/s}$
• B. (a) $1\text{s}$, (b) $8\text{m}$, (c) $8\sqrt{5}\text{m/s}$
• C. (a) $\frac{11}{5}\text{s}$, (b) $\frac{88}{5}\text{m}$, (c) $4\sqrt{5}\text{m/s}$
• D. (a) $1\text{s}$, (b) $8\text{m}$, (c) $4\sqrt{5}\text{m/s}$

Answer 1

The correct option is B (a) $1\text{s}$, (b) $8\text{m}$, (c) $8\sqrt{5}\text{m/s}$

$u_x=10\cos {37}^{\circ }=8\text{m/s},u_y=-10\sin {37}^{\circ }=-6\text{m/s}$

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

b) Range $=u_x\times t=8\times 1=8\text{m}$

c) Speed just before striking the ground $v=\sqrt{v_x^2+v_y^2}=\sqrt{u_x^2+u_y^2+2gh}=\sqrt{(8{)}^2+(6{)}^2+2\times (-10)\times (-11)}=\sqrt{100+2\times 10\times 11}$
$\Rightarrow v=\sqrt{320}\text{m/s}=8\sqrt{5}\text{m/s}$