The height reached in time t by a particle thrown upwared with a speed u is given by
`h=ut-1/2 gt^2`
where g=9.8 m/s^2 is a constant.Find the timetakenin raching the maximum height.

The height reached in time t by a particle thrown upward with a speed u is given by h=ut-1/2 g t^2 where g=9.8 m/s^2 is a constant. Find the time taken in reaching the maximum height.

The height reached in time t by a particle thrown upward with a speed u is given by h=ut-1/2 g t^2 where g=9.8 m/s^2 is a constant. Find the time taken in reaching the maximum height.

The height reached in time t by a particle thrown upward with a speed u is given by h=ut-(1)/(2) "gt"^(2) where g is acceleration due to gravity. Find the time taken in reaching the maximum height.

A stone is thrown up with a speed of 9.8ms^-1 . Find the time taken to reach the maximum height.

A ball is thrown vertically upwards with a velocity of 4.9m/s Find (i) the maximum height reached by the ball (ii) time taken to reach the maximum height

A cricket ball is thrown up with a speed of 19.6 m/s. The maximum height it can reach is (take g= 9.8 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) ).

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) ).

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

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 =