an object is dropped from the top of a cliff 675 meters high. its height above the ground t seconds after it is dropped is 675-4.9t^2. determine its speed 7 seconds after it is dropped, the speed if the object 7 seconds after it's dropped in m/sec

V= 68.6 m/s

Step-by-step explanation:

height above the ground t seconds after it is dropped is 675-4.9t^2

For seven seconds

Height above ground= 675-4.9(7)²

height above ground= 675-240.1

Height above ground= 434.9 m

But it's dropped from a cliff and we need to find the distance traveled from the too of the cliff

Distance= 675-434.9.

Distance= 240.1

To determine it's speed

V²= U² +2gs

V= velocity we are looking for

U= initial velocity= 0

G= acceleration due to gravity= 9.8 m/s²

s= distance covered

V²= U² +2gs

V²= 0+2(9.8)(240.1)

V²= 4705.96

V= 68.6 m/s

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: