Test: Motion in a Plane - 2 - JEE MCQ

velocity at the highest point 

velocity at the starting point 

This is a projectile motion problem. Given that, the range is 100m, we have to find the maximum height that a person can throw a stone.
Range is given as R = u²sin2θ / g where u is the initial velocity, θ is the launch angle and g is the acceleration due to gravity.
Since, R=100m, thus u²sin2θ / g=100 and the acceleration due to gravity g = 9.8.
Thus, it can be written that, u²sin2θ =100g where g=9.8.
It is known to us that, the greatest range is achieved using a 45^∘ launch angle:
u²sin(2×45^∘)=980
⇒ u²sin90^∘= 980

⇒ u²sin⁡90^∘=980
We know that, sin 90^∘=1,
∴ u² = 980
The height is achieved when thrown straight up at 45^∘ angle.
Thus, height H=u²sin²θ / 2g.
Since, u²=980, θ=45^∘ and g=9.8
H=980 / 39.2
⇒ H = 25m

Correct Answer is C.

using the formula:
D = u√2H/g
Where D is distance, u is initial velocity, H is height and g is acceleration due to gravity.
Putting the values,
D= 150√2×1254.4/9.8
D= 150√2508.8/9.8
D= 150√256
D= 150×16
D= 2400m
D= 2.4 km

Range is same for angles of projection θ and 90⁰ - θ

It is clear from the adjoining figure, 

If T is doubled, then R becomes 4 times. 

when a projectile is projected at an angle θ or at angle ( 90^o - θ ) with the horizontal, the horizontal range remains the same. The horizontal range is maximum  when the angle of projection is 45^o

Average velocity = Displacement/Time

Here,  H = maximum height =

R = range = 

nd T = time of flight = 

Angular velocity is always directed perpendicular to the plane of the circular path. Hence, required change in angle = 0^o.

For a particle moving in a circle with constant angular speed, the velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of the circle or is always along the radius of the circle.
Since the tangential vector is perpendicular to radial vector, therefore, velocity vector will be perpendicular to the acceleration vector. Acceleration only causes change in direction of velocity at all instants of time because in no case, acceleration vector is tangent to the circle.

Hence option B is the only false statement.

Uniform straight line motion is a type of linear motion. It works according to Newton's First law of motion i.e. objects that do not experience any net force will continue to move in a straight line with a constant velocity until they are subjected to a net force.

 As both the trains are moving in the opposite direction, the relative speed of each train with respect to each other be, v=10+15=25m/s
Here distance covered by each train = sum of their lengths =50+50=100m
∴ Required time=100/25​=4sec

Lets say the speed of second car be v. Now with respect to car 1, speed of car 2 is v - 30. Now if the cross each other at just after 4 min then the distance traveled by car 2 relative to car 1 is (v -30) x 4/60 and we know the distance travelled relatively is 5km
Thus we get 
(v -30) x 4/60 = 5
V = 300/4 - 30
= 45 km/h

To reach the exact opposite end the parallel component of velocity would nullify the velocity of the river, hence we get 6.sin a = 3 , where a is the angle with perpendicular to river flow.
Thus we get sin a = ½ and cos a =√3/2
Thus the perpendicular component of velocity, 6.cos a =  3√3
Thus time taken is 2 / 3√3