Question

In the war zone, an army tank A is approaching the enemy tank B as shown in figure. A shell is fired from tank A with muzzle velocity $v_0$ at an angle ${37}^0$ to the horizontal at the instant when tank B is 60 m away. Tank B which is moving away with velocity $60m{s}^{-1}$ is hit by shell, then $v_0$ is (g = 10 $m/s^2$)

• A. $25(1+\sqrt{2})m{s}^{-1}$
• B. $50(1+\sqrt{2})m{s}^{-1}$
• C. $80(1+\sqrt{2})m{s}^{-1}$
• D. $75(1+\sqrt{2})m{s}^{-1}$

Answer 1

The correct option is A $25(1+\sqrt{2})m{s}^{-1}$

According to the question we have to use projectile

motion formula to solve

Range = $\frac{2u_x{u}_y}{g};$ Time period = $\frac{2u_y}{g}$

Then, for the situation

$R_A=60+R_B$

$\frac{2(v_0-cos{37}^{\circ }+20)\left(v_{0}sin{37}^{\circ }\right)}{g}=60+\frac{2\left(v_{0}sin{37}^{\circ }\right)}{g}\times 60$

$\frac{2(v_{0}4/5+20)\left(v_0.3/5\right)}{g}=60+\frac{2\left(v_{0}3/5\right)}{g}\times 60$

$\frac{2(4v_0+100)\left(3v_0\right)}{25\times 10}=60+\frac{2\left(3v_0\right)\times 60}{10\times 5}$

$\frac{6v_0(4v_0+100)}{25\times 10}=\frac{60\times 50+120\left(3v_0\right)}{10\times 5}$

$6v_0(4v_0+100)=60\times 50\times 5+120\times 3v_0\times 5$

$24v_0^2+600v_0=15000+1800v_0$

$24v_0^2+600v_0-1800v_0-15000=0$

$24v_0^2-1200v_0-15000=0$

$v_0^2-50v_0-625=0$

Using quadratic formula to solve

$v_0=\frac{50\pm \sqrt{2500+4\times 625}}{2}$

$v_0=\frac{50\pm 50\sqrt{2}}{2}$

$v_0=25(1\pm \sqrt{2})$

Thus, the velocity of the shell fired from

tank A is $25(1+\sqrt{2})m/s$