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, 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

The height h , in feet , of a ball shot upward from a ground level spring gun is described by the formula h=-16t^2+48t . Where t is the time in seconds. What is the maximum height , in feet , reached by the ball ?

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 projected vertically upward with initial velocity of 112 feet per second moves according to the equation s=112t-16t^2 where s is the distance , in feet , from the ground , and t is time , in seconds. What is the maximum height reached by the stone ?

A stone is project vertically upward from the top of a tower of height 60 m. The stone moves according to the law S=(10)t-5t^(2) , where S is in meters and t in second The maximum height , reached by the stone , from the ground , is

A ball is thrown vertically upwards, moves according to the law s=13.8t-4.9t^(2) where s is in metres and t is in seconds. (i) Find the acceleration at t = 1 (ii) Find velocity at t = 1 (iii) Find the maximum height reached by the ball?

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