Question

A ball of mass m is thrown at an angle is 'theta' with the horizontal with an initial velocity u. The change in its momentum during its flight in a time interval of 't' is​

Answer 1

Answer:

The answer is -mgt.

Explanation:

We know that a projectile action is a combination to

horizontal and vertical movement. The vertical part of initial velocity ‘ u ’

always remains steady.

Therefore in this case  u x= u cos o , throughout the

motion. Hence change in momentum in horizontal direction is zero

Now for the vertical movement we know initial velocity

in vertical direction is u y = u sin o.

After time t.

v y = u sin(o)– g t , ( Using equation of motion

v = u + a t )

Thus, Change in momentum in vertical track = m . v y – m . uy 

= m . ( u sin o – g t ) - ( m . u sin o )

= - m g t

Answer 2

mgt

Explanation:

f × t

f = mg

f × t = mg × t

= mgt

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