Astone is throw vertically upward with the speed of 18m/s. a)how far is it moving when it reaches a height of 11 m b) how long is required to reach this height

(a) We can use a simple kinematic equation to find the speed of the stone when it reaches a height of 11 meters. v^2 = (v0)^2 - 2gy v = sqrt{ (18 m/s)^2 - (2)(9.8 m/s^2)(11 m) } v = 10.4 m/s The speed is 10.4 m/s when the stone reaches a height of 11 meters. (b) We can use a simple kinematic equation to find the time it takes for the stone's speed to decelerate to 10.4 m/s. t = (v - v0) / g t = ( 10.4 m/s - 18 m/s ) / (-9.80 m/s^2) t = 0.7755 seconds It takes 0.7755 seconds for the stone to reach a height of 11 meters, where the speed is 10.4 m/s.

The answer is a.  i hope this