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

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step-by-step explanation:

answer:

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step-by-step explanation: lmk if you need a explanation