A stone thrown upwards, has its equation of motion `s=490t-4.9t^2`. The the maximum height reached by it 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, 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 from ground level, has its equation of height h=490t-4.9t^(2) where 'h' is in metres and t is in seconds respectively. What is the maximum height reached by it ?

A stone thrown upwards from ground level, has its equation of height h=490t-4.9t^(2) where 'h' is in metres and t is in seconds respectively. What is the maximum height reached by it ?

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 a speed of 28 m/s. (a) Find the maximum height reached by the stone. (b)Find its velocity one second before it reaches the maximum height.

A stone thrown vertically upwards rises S ft in t seconds where S =112t-16t ^(2) . The maximum height reached by the stone is

A stone thrown vertically upwards satisfies the equations s = 80t –16t^(2) . The time required to reach the maximum height is

A stone projected vertically upward moves according to the law s = 100t-16t^2 . The maximum height reached is