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.
Figure shows position versus time graph of two rabbits running opposite to each other between two trees. Which of the following statements are true.
Distance is equal for both A and B. B is taking lesser time to complete so its speed is higher.
A body is dropped from a tower. It covers 64% distance of its total height in last second. Find out the height of tower [g = 10 ms–²]
\[ 0.36 H= \frac{1}{2}g(t-1)^{2} \]
\[ H= \frac{1}{2}g(t)^{2} \]
Dividing we get
\[ 0.36 = \frac{(t-1)^{2}}{t^{2}} \]
\[ 0.6 = \frac{(t-1)}{t} \]
0.4t =1
t= 2.5 sec
S= (1/2) ×10 ×(2.5)²= 31.25 m
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 boat takes two hours to travel 4 km down and 4 km up the river when the water is still. How much time will the boat take to make the same trip when the river starts flowing at 2 kmph?
Let the boat's speed in still water be x kmph. Given
, we get x = 4 kmph.
With a 2 kmph current:
hours
hours
Total time =
hours or 2 hours 40 minutes.
A taxi leaves the station X for station Y every 10 minutes. Simultaneously, a taxi leaves the station Y also for station X every 10 minutes. The taxis move at the same constant speed and go from X to Y or vice-versa in 2 hours. How many taxis coming from the other side will each taxi meet enroute from Y to X ?
Each taxi takes 2 hours, and taxis leave every 10 minutes.
Taxis already on the route from X to Y:
Taxis that start after from X to Y:
Exclude the simultaneously departing taxi:
Final Answer: 23 taxis.
Two cars A and B are approaching each other head-on with speeds 20 m/s and 10 m/s respectively. When their separation is X then A and
B start braking at 4 m/s² and 2m/s² respectively. Minimum value of X to avoid collision is
Using the stopping distance formula:
For Car A (
m/s,
m/s²):
For Car B (
m/s,
m/s²):
Total minimum separation to avoid collision:
An object is projected horizontally from a 100 m high cliff with a speed of 10 m/s. What will be its velocity 1 second after projection?
An object is projected horizontally from a 45 m high cliff with a speed of 40 m/s. With what speed will it strike the ground? (Take g=10 m/s²)
s
m/s
m/s
Answer: 50 m/s
The relation between times t and distance x is \(t = ax^2 + bx\), where a and b are constants. The acceleration is
Differentiating \(t = ax^2 + bx\) with respect to \(t\) gives \(1 = (2ax + b)v\). Differentiating again with respect to \(t\) gives \(a = \frac{dv}{dt} = -2av^3\).