A particle is thrown with the speed u at an angle α with the horizontal. When the particle makes an angle β with the horizontal, its speed will be :
During Projectile motion horizontal component of velocity remains same so,
\[ u cos\alpha = v cos \beta \]
\[ \frac{u cos\alpha}{cos \beta} = v \]
\[ {u .cos\alpha}.{sec \beta} = v \]
If a body A of mass M is thrown with velocity v at an angle 30° to the horizontal and another body B of same mass is thrown with same speed at an angle of 60° to the horizontal, the ratio of range of A and B will be :
For Complementary angles ranges are equal
Which of the following statements is incorrect :
The statement is wrong because:
If an object moves in a circular path and returns to its starting point, the displacement is 0 (since the initial and final positions are the same), but the path length is the total distance traveled around the circle.
Thus, displacement is not always equal to the path length
The three initial and final positions of a man on the x-axis are given as :-
(i) (–3m, 7m)
(ii) (7m, –3m)
(iii) (–7 m, 3m)
Which pair gives the negative displacement ?
A drunkard is walking along a straight road. he takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after :
### Given:
- The drunkard takes **5 steps forward** (5 meters) and **3 steps backward** (3 meters).
- Each step is **1 meter** and takes **1 second**.
- There is a pit **11 meters** away.
### Correct Solution:
1. In **1 full cycle** (5 steps forward + 3 steps backward), the net distance covered is:
\[
5 - 3 = 2 \text{ meters}.
\]
This cycle takes **8 seconds**.
2. We need to find out how long it takes the drunkard to fall into the pit which is **11 meters** away.
### Step-by-step process:
- After **1 cycle** (8 seconds), the drunkard is at **2 meters**.
- After **2 cycles** (16 seconds), the drunkard is at **4 meters**.
- After **3 cycles** (24 seconds), the drunkard is at **6 meters**.
Now, in the **next (4th) cycle**, the drunkard will walk forward:
- After the first **5 steps forward** (which takes 5 seconds), he will move from **6 meters** to **11 meters**, falling into the pit.
### Total time:
\[
24 \text{ seconds} + 5 \text{ seconds} = 29 \text{ seconds}.
\]
Conclusion:
The drunkard will fall into the pit after **29 seconds**.
A car is moving along a straight line OP as shown in the figure. It moves from O to P in 18 s and returns from P to Q in 6 s. Which of the following statements is not correct regarding the motion of the car :
Total Distance = OP + PQ = 360 + 120 = 480 m
Displacement = OQ= 240 m
Average Speed = 480/24 = 20 m/s
Average Velocity= 240/24= 10 m/s
A car moving along a straight road with speed of 144 km h–¹ is brought to a stop within a distance of 200 m. How long does it take for the car to stop :
To solve this, we'll use the equation of motion:
\[
v^2 = u^2 + 2as
\]
Where:
- \( v = 0 \) (final velocity, since the car comes to rest)
- \( u = 144 \, \text{km/h} = 40 \, \text{m/s} \) (initial velocity)
- \( s = 200 \, \text{m} \) (distance)
- \( a \) is the acceleration (which we'll solve for)
Rearranging the equation:
\[
0 = (40)^2 + 2 \cdot a \cdot 200
\]
\[
0 = 1600 + 400a
\]
\[
a = - \frac{1600}{400} = -4 \, \text{m/s}^2
\]
Now, to find the time (\( t \)):
\[
v = u + at
\]
\[
0 = 40 + (-4)t
\]
\[
t = \frac{40}{4} = 10 \, \text{seconds}
\]
Thus, it takes 10 seconds for the car to stop.
A scooter is going towards east at 10 ms–¹ turns right through an angle of 90°. If the speed of the scooter remains unchanged in taking this turn, the change in the velocity of the scooter is :
Change in a vector when it is rotated by an angle θ is
\[ \Delta V = 2V sin \left( \frac{\theta}{2} \right) \]
\[ \Delta V = 2\times 10 sin \left( \frac{90^{0}}{2} \right)= 20/\sqrt{2}= 10\sqrt{2}= 14.4 m/s \]
Two balls are dropped to the ground from different heights. One ball is dropped 2 s after the other but they both strike the ground at the same time. If the first ball takes 5 s to reach the ground, then the difference in initial heights is: (Take g = 10 ms–²)
1. Time taken by each ball:
- First ball: \( t_1 = 5 \, \text{s} \)
- Second ball: \( t_2 = 3 \, \text{s} \) (dropped 2 s later)
2. Heights:
- Height of the first ball:
\[
h_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \cdot 10 \cdot (5)^2 = 125 \, \text{m}
\]
- Height of the second ball:
\[
h_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} \cdot 10 \cdot (3)^2 = 45 \, \text{m}
\]
3. Difference in heights:
\[
\Delta h = h_1 - h_2 = 125 - 45 = 80 \, \text{m}
\]
The difference in initial heights is 80 meters.
A river flows from east to west with a speed of 5 m/min. A man on south bank of river, capable of swimming at the rate of 10 m/min in still water, wants to swim across the river in shortest time; he should swim :
To cross the river in the shortest time, the man should swim directly perpendicular to the riverbank, regardless of the river's flow. This way, his entire swimming speed is used to cross the river.
Given:
- Speed of the river (east to west) = 5 m/min
- Speed of the swimmer in still water = 10 m/min
The swimmer should aim **directly north** to minimize the crossing time. The river's current will cause him to drift downstream, but this direction ensures the shortest crossing time.
Thus, the man should swim:
\[
{\text{directly north}}
\]