Two guns on a battleship simultaneously fires two shells with same speed at enemy ships. If the shells follow the parabolic trajectories as shown, which ship will get hit first ?
As A is having more maximum height it will have more vertical speed uy. So A will take more time than A.
So, ship B will be hit first.
A particle is fired with velocity u making an angle θ with the horizontal. What is the change in velocity when it is at the highest point :
At the highest point of its trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged.
- Initial velocity: \( u \)
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)
At the highest point:
- Vertical velocity: \( u_y = 0 \)
- Horizontal velocity: \( u_x = u \cos \theta \)
The change in velocity is only in the vertical direction, from \( u \sin \theta \) to 0.
So, the change in velocity is:
\[
\Delta v = u \sin \theta
\]
Thus, the change in velocity is \( u \sin \theta \).
A particle is projected from horizontal ground with speed \(50\text{ ms}^{-1}\) at \(53^\circ\) with horizontal. Find time after which velocity of particle will be at \(45^\circ\) with horizontal for the second time.
Horizontal speed is \(u_x = u\cos(53^\circ) = 30\text{ m/s}\), and vertical speed is \(u_y = u\sin(53^\circ) = 40\text{ m/s}\). At \(45^\circ\) on the way down, \(v_y = -v_x = -30\text{ m/s}\). Using \(v_y = u_y - gt\), we get \(-30 = 40 - 10t\), which yields \(t = 7\text{ s}\).
A particle is moving in vertical plane (x-y plane) such that its trajectory is given by the equation, \(y = x – \frac{x^2}{80}\), where \(x\) & \(y\) are in metre. For this particle match column-I with column-II and tick the correct option.
**Column I**:
A. Horizontal range (in m)
B. Angle of projection with horizontal (degree)
C. Maximum height gained by particle (m)
D. Speed of projection (\(\text{m s}^{-1}\))
**Column II**:
(P) 45
(Q) 80
(R) 20
(S) \(20\sqrt{2}\)
Comparing the trajectory to \(y = xtan\theta - \frac{gx^2}{2u^2cos^2\theta}\), we find \(tan\theta = 1⇒ \theta = 45^\circ\). Range at \(y = 0\) is \(x = 80\text{ m}\). Maximum height is \(H = \frac{Rtan\theta}{4} = 20\text{ m}\), and the initial speed \(u = 20\sqrt{2}\text{ m/s}\). Hence, A-Q, B-P, C-R, D-S.