Test: Motion in a Plane- 2 - NEET MCQ

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 v² - u² = 2gh
0² - 20² = 2 * (- 9.8) * h
- 400 = -19.6 h
h = - 400 /(-19.6) = 20.408 m (approximately)

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| = √(7² + 7²) = 7√2

A body start from rest so, u = 0 
Let total distance covered = s
Let a body moves with accerlation = a 
s = 1/2 × a× t²
s = 1/2× a × 4²
s = 8a
The one-fourth of total distance, s'  = 8a/4
= 2a
s' = 1/2 × a × t²
2a / a = 1/2 × t²
4 = t²
t = 2 seconds

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/s² in South West

Time taken by first drop to cover 5 cm (u = 0), is: 
h = 1/2 gt²
5 = 1/2 x 10 x t²
t = 1 sec
Hence, the interval is 0.5 sec for each drop.
Now distance fallen by second drop in 0.5 sec
h₁ = 1/2 gt²
= 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