The maximum height is reached is `5s` by a stone thrown vertically upwards and moving under the equation `10s=10ut-49t^(2)`, where `s` is in metre and `t` is in second. The value of `u` is

The maximum height is reached in 5s by a stone thrown vertically upwards and moving under the equation 10s=10ut-49t^(2) , where s is in metre and t is in second. The value of u is: a) 4.9 m/sec b) 49 m/sec c) 98 m/sec d)None of these

A stone thrown vertically upward satisfies the equation s= 64 t -16t^(2) where is in meter and t is second .What is the time required to reach the maximum height ?

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

Find the maximum height reached by a particle which is thrown vertically upwards with the velocity of 20 m//s (take g= 10 m//s^(2) ).

Find the maximum height reached by a particle which is thrown vertically upwards with the velocity of 20 m//s (take g= 10 m//s^(2) ).

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

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