Question

An object of mass 20kg is projected from ground with speed 30m/s at an angle of 30 degree with horizontal.The rate of change of momentum per second of object at maximum height will be [g=9.8]

Answer 1

initial speed of projectile 30m/s at an angle 30° with horizontal .
so, initial velocity , $\bf{\vec{u}=30cos30\hat{i}+30sin30\hat{j}}$
$\bf{\vec{u}=15\sqrt{3}\hat{i}+15\hat{j}}$

so, initial momentum of object , $P_i$
= mu
= 20 × (15√3i + 15j)
= 300√3 i + 300 j .......(i)

velocity at maximum height ,
v = 30cos30° i [ as you know only horizontal component will be at maximum height]
so, final momentum of object , $P_f$
= mv = 20 × 30cos30° i= 300√3 i....(ii)

now , time taken to reach maximum height = half of time of flight = usin30°/g
= 30sin30°/10 = 1.5 sec

now , rate of change of momentum = (Pf - Pi)/t
= (300√3i - 300√3i - 300j)/1.5
= -200j

hence, magnitude of rate of change of momentum = 200 kgm/s²

Answer 2

Answer:

answer is 200

Explanation:

hope this helps you