Arock is thrown upward with a velocity of 14 meters per second from the top of a 20 meter high cliff, and it misses the cliff on the way back down. when will the rock be 12 meters from the water, below? round your answer to two decimal places.

3.35 s after the rock is thrown, it will be 12 m above the water

Step-by-step 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.

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo

the solution is x=1

step-by-step explanation:

we are given our   equation as

let's assume

left side function is

right side function is

now, we can draw graph

and then we can get intersection point

the intersection point will be solution

so, the solution is

x=1