Answered

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The Function approximate the velocity (in feet per second) of an object dropped from a height of x feet
right before it hits the ground on Earth and on Mars.
Earth: E(X) = 8V
Mars: M(x) = 0.6(E(x))
What is the velocity of an object dropped from a height og 25 feet right before it hits the ground on
Mars?

Sagot :

Answer:

The answer to the question is

It take for the golf ball hit on Mars to reach the ground 8.284 s longer.

Step-by-step explanation:

To Solve the question, we note that the equation of motion is given to us as

h = -1.9·t² + 26·t

Therefore the height

We compare the above equation with the equation of of motion

S = ut - 0.5×g×t² where u = initial velocity = 26 m/s and  - 0.5×g = -1.9

Therefore the acceleration due to gravity om mars = -1.9/-0.5 = 3.8 m/s²

We now have the time taken to maximum height given by

v = u - g·t where v = 0 at maximum height

We therefore have u = g·t → 26 = 3.8×t or t = 26/3.8 = 6.84 seconds

The time taken for ball to hit the ground is 2 × the time to reach maximum height = 2 × 6.84 = 13.684 s

The time longer it takes for the ball on Mars to complete the path up to maximum height and down again is 13.684 s - 5.4 s = 8.284 s.

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