prove that if a body is thrown vertically upward , the time of ascent is equal to the time of descent.

Ascent:

Time of ascent be t seconds and the object be thrown with an initial velocity be u m/s.

Descent:

Time of descent be t' seconds and the object final velocity be v m/s. The distance covered will be the same as when the object is thrown upwards.

From equation (1) and (2) we can see that u = v. 

t' = t, that is, time of ascent is equal to time of descent.

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hmm qstn frm gravition or wht?

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During acsent:

g=9.8 , t1=time taken for ascent , u=initial velocity , v=final velocity=0

Therefore,

t1= (v-u)/-g= (0-u)/g = u/g

During descent,

g=9.8 , u=initial velocity=0 , v=final velocity, t2=time taken for the descent

Therefore,

t2= (v-u)/g= (v-o)/g= v/g

The velocity with which the body is thrown = The velocity with which the body hits the ground. (experimented truth)

Therefore,

u/g=v/g

Thus, t1=t2

or, the time of ascent = the time of descent (PROVED)

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