Question

If two bodies A and B are projected upwards such that the velocity of A is double the velocity of B, then the height to which the body A will rise will be two times the height to which the body B will rise

Answer 1

Answer:

Answer:

Maximum height attained by a projectile

h =

2gR−v

2

v

2

R

. (i)

Velocity of body = half the escape velocity

v=

2

v

e

Or v=

2

2

gR

→v

2

=

4

2gR

Or v

2

=

2

gR

Now, putting value of v2 in Eq. (i),we get

h=

2gR−

2

gR

2

gR

R

=

3gR/2

gR

2

/R

Or h=

3

R

Answer 2

Answer:

Body A will rise 4 times higher than body B.

Explanation:

Let, U be the velocity of A and

     V be the velocity of B

Since, velocity of A is double the velocity of B.

therefore, U = 2V         ------- 1

let, h( A ) be the height of A and

     h( B ) be the height of B

so, 0^2 = U^2 + 2( -g ) h( A )

    h( A ) =U^2/2g         -------- 2

    0^2 = V^2 + 2( -g ) h( B )

    h( B ) = V^2/2g        --------- 3

Now, from equation  1 and 2

        h( A ) = ( 2V )^2/2g

        h( A ) = 4V^2/2g

        h( A ) = 4 h( B )        from equation 3

Hence, body A will rise 4 times higher than body B