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A stone is thrown upwards with initial velocity of 20 m s^{1}, the height that stone will reach would be:
Initial velocity = u = 20 m/s
Final velocity = v = 0 ( at maximum height , v = 0 )
Acceleration due to gravity(g) in this case, is taken as negative.
This is because when the direction of motion of object is opposite to "g" , then value of g is taken as ve.
Hence, g = 9.8 m/s²
Using v^{2}  u^{2} = 2gh
0^{2}  20^{2} = 2 * ( 9.8) * h
 400 = 19.6 h
h =  400 /(19.6) = 20.408 m (approximately)
A particle has an initial velocity of 3i + 4j and an acceleration of 0.4i + 0.3j. Its speed after 10s is:
Use equation of kinematics: v = u + at
Initial velocity, u = 3i + 4j
Acceleration, a = 0.4i + 0.3j
Time, t = 10s
v = (3i + 4j) + 10(0.4i + 0.3j)
v = (3i + 4j) + (4i + 3j)
v = 7i + 7j
v = √(7^{2 }+ 7^{2}) = 7√2
A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover onefourth distance starting from the rest at the top?
A body start from rest so, u = 0
Let total distance covered = s
Let a body moves with accerlation = a
s = 1/2 × a× t^{2}
s = 1/2× a × 4^{2}
s = 8a
The onefourth of total distance, s' = 8a/4
= 2a
s' = 1/2 × a × t^{2}
2a / a = 1/2 × t^{2}
4 = t^{2}
t = 2 seconds
A car travelling at 36km/h^{1} due North turns West in 5 seconds and maintains the same speed. What is the acceleration of the car?
Initial velocity of car = 36 km/hr due north
Final velocity of car = 36 km/hr due west
The magnitude of change in velocity:
(since velocity is a vector, so direction has to be taken into account)
Acceleration = (Change in Velocity) / Time
= [36(√2) * 5/18] / 5
= 2(√2) m/s^{2} in South West
Water drops fall at regular intervals from a tap that is 5m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant?
Time taken by first drop to cover 5 cm (u = 0), is:
h = 1/2 gt^{2}
5 = 1/2 x 10 x t^{2}
t = 1 sec
Hence, the interval is 0.5 sec for each drop.
Now distance fallen by second drop in 0.5 sec
h_{1} = 1/2 gt^{2}
= 1/2 x 10 x (0.5)^{2}
= 5 x 0.25
= 1.25 m
Height above the ground ( of 2^{nd} drop) = 5  1.25 = 3.75 m
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