A stone, vertically thrown upward is moving in a line. Its equation of motion is `s=294t-49t^(2)`, then the maximum height that the stone reaches is

A stone thrown upwards, has its equation of motion s=490 t-(4.9)t^(2) . Then the maximum height reached by it is

A stone thrown upwards, has its equation of motion s=490t-4.9t^2 . The the maximum height reached by it is

A stone is thrown in upward direction. Its equation is S=80 t-16 t^(2) . The time to attain maximum height is …………. second.

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g= 10 m//s^2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

A stone is thrown vertically upward with an initial velocity of 40ms^-1 . Taking g = 10ms^-2 , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone ?

A stone is thrown verticaly upward with an initial velocity of 40m//s . Taking g=10m//s^(2) , find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?