A point moves in $x-y$ plane as per $x=kt,$ $y = kt\left( {1 - \alpha t} \right)$ where $k\,\& \,\alpha \,$ are $+ve$ constants. The equation of trajectory is
$y = x - \frac{{\alpha {x^2}}}{k}$
$y = x + \frac{{\alpha {x^2}}}{k}$
$x = y - \frac{{\alpha {y^2}}}{k}$
$x = y - \frac{{\alpha {y^2}}}{k}$
A particle is moving along a curve. Then
$Assertion$ : If a body is thrown upwards, the distance covered by it in the last second of upward motion is about $5\, m$ irrespective of its initial speed
$Reason$ : The distance covered in the last second of upward motion is equal to that covered in the first second of downward motion when the particle is dropped.
A particle has an initial velocity of ($3\hat i + 4\hat j)\;ms^{-1}$ and an acceleration of $(0.4\hat i + 0.3\hat j)\;ms^{-1}$ Its speed after $10\;s$ is:
A projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$, the range of the projectile is $d$. If at the highest point in its trajectory, the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$, then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .
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 ?