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david drops a soccer ball off a building. the building is 75 meters tall. (a) what is the height of the ball to the nearest tenth of a meter exactly 1 seconds after he drops the ball? (b) how many seconds, after the ball is released, will it hit the ground?

(a) 70.1 m

The height of the ball at time t is described by the following equation:

(1)

where

is the initial height of the ball

is the acceleration due to gravity

t is the time

Since we want to know the heigth of the ball after 1 second, we just need to substitutite t=1 s into the formula, and we find:

(b) 3.9 s

To find the time it takes for the ball to reach the ground, we have to find the time t at which the height of the ball h(t) is zero: h(t) =0. Using eq.(1), this means:

And re-arranging we find:

x= -3

step-by-step explanation:

step 1: simplify brackets

4 - x - 1 = 6

step 2: simpilfy

3 - x = 6

step 3: subtract 3 from both sides

-x = 6 - 3

step 4: simpilfy 6 - 3

-x = 3

step 5: multiply both sides by -1

x = -3

step-by-step explanation: