A fighter plane is flying horizontally at an altitude of $1.5\, km$ with speed $720\, km/h$. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target ?
Suppose fighter plane, when it is at $P$ drops a bomb to hit a target$ T$.
Speed of the plane $=720 \mathrm{~km} / \mathrm{h}$
$=720 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=200 \mathrm{~m} / \mathrm{s}$
Height of the plane $\left(\mathrm{P}^{\prime} \mathrm{T}\right)=1.5 \mathrm{~km}=1500 \mathrm{~m}$
If bomb hits the target after time $t$, then horizontal distance travelled by the bomb, $\mathrm{PP}^{\prime}=u \times t=200 t$
Vertical distance travelled by the bomb,
$\mathrm{P}^{\prime} \mathrm{T}=\frac{1}{2} g t^{2}$
$\therefore \quad 1500=\frac{1}{2} \times 9.8 t^{2}$
$\therefore \quad t^{2}=\frac{1500}{4.9}$
$\therefore \quad t=\sqrt{\frac{1500}{4.9}}=17.49 \mathrm{~s}$
Using value of $t$ in Eq.(i)
$\mathrm{PP}^{\prime}=200 \times 17.49 \mathrm{~m}$
Now, $\tan \theta=\frac{\mathrm{P}^{\prime} \mathrm{T}}{\mathrm{P}^{\prime} \mathrm{P}}=\frac{1500}{200 \times 17.49}$
$=49287=\tan 23^{\circ} 12^{\prime}$
The displacement $x$ of a particle depend on time $t$ as $x = \alpha {t^{^2}} - \beta {t^3}$
The co-ordinates of a particle moving in $x-y$ plane are given by : $\mathrm{x}=2+4 \mathrm{t}, \mathrm{y}=3 \mathrm{t}+8 \mathrm{t}^2 .$ The motion of the particle is :
The initial velocity of a projectile is $\vec u = (4\hat i + 3\hat j)\,m/s$ it is moving with uniform acceleration $\vec a = (0.4\hat i + 0.3\hat j)\, m/s^2$ The magnitude of its velocity after $10\,s$ is.........$m/s$
A particle moves in the $xy$ plane with a constant acceleration $'g'$ in the negative $y$-direction. Its equation of motion is $y = ax-bx^2$, where $a$ and $b$ are constants. Which of the following are correct?