Question

DERIVE THE EQUATIONS OF MOTION FOR A PARTICLE
PROTECTED VERTICALLY
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Answer 1

Answer:

’Equations of motion for a particle falling vertically downward from certain height. Consider an object of mass m falling from a height h. Assume there is no air resistance. For convenience, let us choose the downward direction as positive y – axis as shown in the figure. The object experiences acceleration ‘g’ due to gravity which is constant near the surface of the Earth. We can use kinematic equations to explain its motion. We have The acceleration a⃗ a→= gi^i^ By comparing the components, we get, Equations of motion for a particle thrown vertically upwards, ax = 0, ax = 0, ay = g Let us take for simplicity, ay = a = g If the particle is thrown with initial velocity ‘u’ downward which is in negative y – axis, then velocity and position at of the particle any time t is given by v = u + gt v = ut + – gt2 The square of the speed of the particle when it is at a distance y from the hill – top is V2 = u2 + 2gy Suppose the particle starts from rest. Then u = 0 Then the velocity v2, the position of the particle and v at any time t are given by (for a point y from the hill – top) v = gt …………(i) y = 1212 – gt2 …………(ii) v2 = 2gy …………(iii) The time (t = T) taken by the particle to reach the ground (for which y = h), is given by using equation (ii), h = 1212– gT2 …………(iv) T = 2hg−−√2hg …………(v) The equation (iv) implies that greater the height (h), particle takes more time (T) to reach the ground. For lesser height (h), it takes lesser time to reach the ground. The speed of the particle when it reaches the ground (y = h) can be found using equation (iii), we get, vgroundvground = 2gh−−−√2gh …………(vi) The above equation implies that the body falling from greater height (h) will have higher velocity when it reaches the ground. The motion of a body falling towards the Earth from a small altitude (h<