Answer:
The maximum height of the rock is 14.2 m
Explanation:
The equations that describe the height and velocity of the rock are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the object at time t
y0 = initial height
t = time
g = acceleration due to gravity (-9.8 m/s² if upward is positive)
v = velocity of the object at time t
We know that at t = 3.40 s, the rock is in your hand again. Then, if we place the origin of the frame of reference at your hand, the position of the rock at 3.40 s is 0 m. Using the equation of the position, we can calculate the initial velocity that we will need to obtain the max-height.
y = y0 + v0 · t + 1/2 · g · t²
0 = v0 · 3.40 s - 1/2 · 9.8 m/s² · (3.40 s)²
(1/2 · 9.8 m/s² · (3.40 s)² ) / 3.40 s = v0
v0 = 16.7 m/s
At max-height, the velocity of the rock is 0. Then, using the equation of velocity we can calculate the time it takes the rock to reach the max-height. With that time, we can calculate the maximum height.
v = v0 + g · t (at max-height, v = 0)
0 = 16.7 m/s - 9.8 m/s² · t
- 16.7 m/s / - 9.8 m/s² = t
t = 1.70 s
Now, using this time in the equation of height:
y = y0 + v0 · t + 1/2 · g · t²
y = 0 m + 16.7 m/s · 1.70 s - 1/2 · 9.8 m/s² · (1.70 s)²
y = 14.2 m
The maximum height of the rock is 14.2 m
