Answer:
3.35 s after the rock is thrown, it will be 12 m above the water
Explanation:
Hi there!
The height of the rock can be calculated using the following equation:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height of the rock at time t.
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
Let´s place the origin of the frame of reference on the water so that y0 = 20 m.
Using the equation of height, we can obtain the time at which the rock is at a height of 12 m above the water:
y = y0 + v0 · t + 1/2 · g · t²
12 m = 20 m + 14 m/s · t + 1/2 · (-9.8 m/s²) · t²
subtract 12 to both sides of the equation:
0 = -12 + 20 m + 14 m/s · t - 4.9 m/s² · t²
0 = 8 m + 14 m/s · t - 4.9 m/s² · t²
Let´s solve the quadratic equation using the quadratic formula:
a = -4.9
b = 14
c = 8
t = [-b ± √(b² - 4ac)] / 2a
t = 3.35 s and t = -0.49 s
Since time can´t be negative, the rock will be 12 m above the water 3.35 s after it is thrown.