A stone is thrown vertically upward with a speed of 28 m/s. (a) Find the maximum height reached by the stone. (b)Find its velocity one second before it reaches the maximum height. (c)Does the answer of part b change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s?

Answer

C, D

Explanation

Given `u=28 m/s, v=0, a =-g=-9.8 m/s^2, S_2 = (v^2-u^2)/(2a), =(0^2-28^2)/(2(-9.8)=2.85, t'=2.85-1=1=1.85, v'=u+at'=28-(9.8)(1.85), (28-18.13=9.87m/s` Hence the velocity is 9.87m/s. c. No it will not change. As, after one second velocity becomes zero for any initial velocity and deceleration is `g= 9.8 m/s^2` remains same. for initial velocity more than 28 m/s maximum height increases.