A stone is thrown vertically upward with a speed of 28 m/s . Determine the maximum height reached by the stone, its velocity one second before it reaches the maximum height, and whether this velocity changes if the initial speed is more than 28 m/s , such as 40 m/s or 80 m/s .

Step-by-step solution

Using the third equation of motion, v^2 = u^2 - 2gH . At maximum height, v = 0 . Thus, H = u^2 2g . Substituting u = 28 m/s and g = 9.8 m/s ^2 : H = 28^2 2 9.8 = 784 19.6 = 40 m . Let T be the time taken to reach maximum height. At maximum height, v = 0 , so 0 = u - gT T = u g . The velocity one second before reaching maximum height is at time t = T - 1 . Using v' = u - gt : v' = u - g(T - 1) = u - g ( u g - 1 ) = u - u + g = g = 9.8 m/s . Since v' depends only on g and is independent of the initial velocity u , th