Given:
Initial velocity with which the stone is thrown vertically upwards, u = 28 m/s
When the stone reaches the ground, its final velocity (v) is 0.
Also,
a = g = −9.8 m/s2 (Acceleration due to gravity)
(a) Maximum height can be found using the equation of motion.
Thus, we have:
v2 − u2 = 2as
On putting respective values, we get:
(b) Total time taken by the stone to reach the maximum height:
As per the question, we need to find the velocity of the stone one second before it reaches the maximum height.
t' = 2.85 − 1 = 1.85 s
Again, using the equation of motion, we get:
v' = u + at' = 28 − 9.8 × 1.85
⇒ v' = 28 − 18.13 = 9.87 m/s
Hence, the velocity is 9.87 m/s
(c) No answer of part (b) will not change, as after one second, the velocity becomes zero for any initial velocity and acceleration (a = − 9.8 m/s2) remains the same. For any initial velocity more than 28 m/s, only the maximum height increases.