Answer:
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:
[tex]y= -\frac{1}{2} gt^{2}[/tex]
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][tex]s^{-2}[/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[tex]s^{-2}[/tex]
[tex] y = -\frac{1}{2} gt^{2}[/tex]
[tex] -5000 = - \frac{1}{2} 32.18 × t^{2}[/tex]
[tex] t^{2} = \frac{5000×2}{32.18}[/tex]
[tex] t^{2} = 311.13876[/tex]
[tex] t = 17.6s [/tex]
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
