The height (in meters) at any time t (in seconds) of a ball thrown vertically varies according to equation `h(t)=-16t^(2)+256t`. How long after in seconds the ball reaches the hightest point

A ball is thrown vertically upwards which satisfies the equation S=80t-16t^2 . Find 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 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 ?

The displacement (in metre) of a particle varies with time (in second) according to the equation y =- (2)/(3) t ^(2) + 16 t +2. How long does the particle take to come to rest ?

A stone projected vertically upward moves according to the law s=48t-16t^2 . The time taken by the stone to reach the point of projection is

When a ball is thrown straight up at an initial velocity of 54 feet per second. The height of the ball t seconds after it is thrown is given by the function h(t)=54t-12t^(2) . How many seconds after the ball is thrown will it return to the ground?

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 is thrown vertically up and the height s reached in time t is given by s=80t-16t^2 . The stone reaches the maximum height in time t =

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