Question

The velocity of a body which has fallen freely under gravity varies as g^ph^q where g is the acceleration due to
gravity at the place and h is the height through which the body has fallen. Determine the value of p and q.​

Answer 1

answer : p = 1/2 and q = 1/2

explanation : A body is falling freely under gravity. so, initial velocity of body will be zero. let v is the velocity of body when it falls h from the initial position.

using formula, v² = u² + 2as

here, a = -g , u = 0 and s = -h

then, v² = 0 + 2(-g)(-h)

or, v² = 2gh

or, v = √(2gh) = √2(gh)½

here it is clear that velocity of a body has fallen freely under gravity varies as (gh)½.

on comparing with g^ph^q ,

we get p = 1/2 and q = 1/2

Answer 2

Answer:

Explanation:

Initial velocity of body = 0. ( since it is falling freely under the gravity)

Let the velocity of body when it falls h from the initial position be = V

Variation = g^ph^q

Thus, using the second equation of motion -

= v² = u² + 2as

where, a = -g , u = 0 and s = -h

On substituting the values -

= v² = 0 + 2(-g)(-h)

or, v² = 2gh

or, v = √(2gh) = √2(gh)½

Thus,  the velocity of a body has fallen freely under gravity varies as (gh)½.

Hence, on comparing with g^ph^q , we will get -

p = 1/2 and q = 1/2