Assume you drop a rock from a cliff which is 5000 feet above the ground. the distance in feet, y, of the rock below the cliff at the time t seconds after it was dropped is given by the equation
y=-1/2gt^2

Step-by-step explanation:

Given that height of cliff H = 5000 ft

You drop rock from cliff and the distance of rock below the cliff at time,t is given by:

Q1. How far below the cliff is the rock 7 seconds after it dropped ?

Ans. Here, t=7 seconds

By using formula provide, g=32.18 ft{tex]

y=- \frac{1}{2} × 32.17 × (7)^{2}[/tex] = 788.165 ‬ft

Note: Negative sign shows the direction and 240.345 is magnitude

Therefore, Rock is 788.165‬ ft from Cliff

Q2. How far is the rock from the bottom 7 seconds after it dropped ?

By using formula provide,

Ans. After 7 seconds, Rock is 788.165‬ ft from Cliff

Therefore, Rock is 5000-788.165‬ = 4,211.835 ft from bottom

Q3. When will rock hit the bottom?

Ans. Height of cliff H= 5000 ft

Using formula provided, g=32.18 ft

Q4. The Velocity of the rock at impact?

Ans. Velocity is given as V= g×t

By using formula, We get

V= 32.14 × Total time to reach ground

V= 32.14 × 17.6 = 565.66 ft/s

c.

step-by-step explanation:

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .