Answer is
9.773m/s^2
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Given,
h=8848m
The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.
g′=g(1 − 2h/h)
=9.8(1 - 6400000/17696)
=9.8(1 − 0.00276)
9.8×0.99724
=9.773m/s^2
Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2
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hope this helps :)
Answer:252 miles
Explanation:
Given
During his way to mountain it took 7 hr to drive
and during his return trip it took 4 hr to return
Let x be the distance between home and mountain
average speed for return is 27 miles per hour faster than his former trip
let v be the speed on his way to mountain thus v+27 is his return speed
thus
----1
for return trip
-----2
divide 1 & 2




thus 
Answer:

Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives

For the y-direction gives

Combining both equation yields the y_component of the final velocity

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.
