Question

suppose a stone thrown vertically upward with initial velocity u reaches a height h before coming down.show that the time taken to be go up is same as the time taken to be come down?

Answer 1

Given that a body is thrown with initial velocity

u in vertically up direction….say along y-axis and it reaches a height h and returns to the thrower’s position.

suppose it takes t1 seconds to reach height h and it returns maening thereby that it has spent all its K E and final velocity is zero at the time t1.

final vel. = 0 = initial vel(u) - deceleration due to gravity(g) x t1

t1 = u / g

while returning path through the same length h it starts with initial velocity zero , however through out the path acceleration due to gravity is giving it the velocity each second.

if the downward travel time is t2 then also the final velocity say u’ as the deceleration has turned into acceleration

so we can write u’ = 0 + gt2

but taking conservation of energy into consideration

touch down kinetic energy = potential energy at height h

and K.E. (start) = P.E. at h =mgh

(1/2) m . u’^2 = mgh = (1/2) m. u^2

therefore u’ = u

so t2= u’/g = u/g = t1

therefore the time of ascending must be equal to time of descending.


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