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 body dropped from the top of a tower clears 7/16th of the total height of the tower in its last second of flight. The time taken by the body to reach the ground is :
According to Galileo's Ratio of Odd Number distances travelled is consecution interval of 1 sec is in the ratio of 1:3:5:7 so in 4th second distance travelled is 7/16th of total journey.
so, total time = 4 sec.
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 \( u \sin \theta \).
Two cars are moving along a straight line in opposite direction with the same speed 10 m/s. The relative velocity of two cars w.r.t. each other is:
When objects move in opposite directions relative velocity= v1+ v2= 20 m/s
Two objects A and B are moving with speeds 10 m/s and 20 m/s respectively in the same direction. The relative velocity of A w.r.t. B is
When moving in same direction relative velocity = V 1 -V 2= 10-20 = -10 m/s
Two objects A and B are moving with speeds 10 m/s and 20 m/s respectively in the same direction. The relative velocity of B w.r.t. A is
Velocity of A w.r.t B = V A - V B
Velocity of B w.r.t A = V B - V A
The relative velocity of two objects A and B is 10 m/s. If the velocity of object A is 40 m/s then the velocity with which B is moving is (assume both objects are moving in same direction)
Relative Velocity = V A - V B
10= 40 - V B
V B = 30 m/s
* Is 50 m/s a possible answer ?
Two trains A and B are moving in a straight line in the same direction with speeds of 54 km/h and 15 m/s respectively. The relative velocity of one train w.r.t. other is
As, 54 km/hr = 15 m/s both have same velocity in same direction their relative speed/velocity is zero
Two objects A and B are moving in same direction with same speed of 20 m/s each then, which of the following position-time graphs correctly represents two moving objects A and B
As both A and B have same velocity their relative speed is zero. so both have same slope in position time graph.
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.