Home
Class 11
PHYSICS
A body is lauched from the earth's surfa...

A body is lauched from the earth's surface a an angle `alpha=30^(@)` to the horizontal at a speed `v_(0)=sqrt((1.5 GM)/R)`. Neglecting air resistance and earth's rotation,
(i)find the height to which the body will rise.
(ii) find radius of curvature at top most point.

A

1.323 R

B

2.323 R

C

3.323 R

D

4.323 R

Text Solution

Verified by Experts

The correct Answer is:
B
Promotional Banner

Similar Questions

Explore conceptually related problems

An object is projected from the earth's surface with escape velocity at 30^(@) with horizontal. What is the angle made by the velocity with horizontal when the object reaches a height 2R from the earth's surface ? R is the radius of the earth. Horizontal can be considered as a line parallel to the tangent at the earth's surface just below the object .

Two bodies were thrown simultaneously from the same point, one straight up, and the other at an angle of theta = 30^@ to the horizontal. The initial velocity of each body is 20 ms^(-1) . Neglecting air resistance, the distance between the bodies at t = 1.2 later is

A projectile of mass m is fired from the surface of the earth at an angle alpha = 60^(@) from the vertical. The initial speed upsilon_(0) is equal to sqrt((GM_(e))/(R_(e)) . How high does the projectile rise ? Neglect air resistance and the earth's rotation.

A body is thrown with the velocity v_0 at an angle of theta to the horizontal. Determine v_0 "in" m s^-1 if the maximum height attained by the body is 5 m and at the highest point of its trajectory the radius of curvature is r = 3 m . Neglect air resistance. [Use sqrt(80)= 9] .

A body is project at t= 0 with a velocity 10 ms^-1 at an angle of 60 ^(@) with the horizontal .The radius of curvature of its trajectory at t=1s is R. Neglecting air resitance and taking acceleration due to gravity g= 10 ms^-2 , the value of R is :

A missile is launched at an angle of 60^(@) to the vertical with a velocity sqrt( 0.75gR) from the surface of the earth ( R is the radius of the earth). Find the maximum height from the surface of earth. (Neglect air resistance and rotating of earth).

A body is thrown with the velocity v_0 at an angle of theta to the horizon. Determine v_0 "in" m s^-1 if the maximum height attained by the body is 5 m and at the highest point of its trajectory the radius of curvature is r = 3 m . Neglect air resistance. [Use sqrt(80) as 9] .

A rocket is launched vertical from the surface of the earth of radius R with an initial speed v . If atmospheric resistance is neglected, then maximum height attained by the rocket is

A body is projected vertically upwards with speed sqrt(g R_(e )) from the surface of earth. What is the maximum height attained by the body ? [ R_(e ) is the radius of earth, g is acceleration due to gravity]

A ball of mass m is fired vertically upwards from the surface of the earth with velocity nv_(e) , where v_(e) is the escape velocity and nlt1 . Neglecting air resistance, to what height will the ball rise? (Take radius of the earth=R):-