Answer:
(i) 6 s
(ii) 45 m
(iii) 2 s
Explanation:
The stone has constant acceleration, so we can model its motion using kinematic equations also known as SUVAT equations. In this case, we will use the two equations:
s = ut + ½ at²
v² = u² + 2as
where
(i) When the stone returns to the ground, its displacement is s = 0. Given that the initial velocity is u = 30, and the acceleration is a = -10, we can use the first equation to solve for time.
s = ut + ½ at²
0 = 30t + ½ (-10)t²
0 = 30t − 5t²
0 = -5t (t − 6)
t = 0 or 6
It takes 6 seconds for the stone to return to the ground.
(ii) When the stone reaches its maximum height, its velocity is v = 0. Given the same initial velocity and acceleration from before, we can use the second equation to solve for displacement.
v² = u² + 2as
0² = (30)² + 2(-10)s
0 = 900 − 20s
s = 45
The stone reaches a maximum height of 45 meters.
(iii) Use the first equation to find the times when the stone has a displacement of s = 40.
s = ut + ½ at²
40 = 30t + ½ (-10)t²
40 = 30t − 5t²
5t² − 30t + 40 = 0
t² − 6t + 8 = 0
(t − 2) (t − 4) = 0
t = 2 or 4
The stone is 40 meters above the ground at 2 seconds and 4 seconds. Therefore, it has a height of more than 40 meters for 2 seconds.