A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
1 answer:
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Answer:
The depth is 5.15 m.
Explanation:
Lets take the depth of the pool = h m
The atmospheric pressure ,P = 101235 N/m²
The area of the top = A m²
The area of the bottom = a m²
Given that A= 1.5 a
The force on the top of the pool = P A
The total pressure on the bottom = P + ρ g h
ρ =Density of the water = 1000 kg/m³
The total pressure at the bottom of the pool = (P + ρ g h) a
The bottom and the top force is same
(P + ρ g h) a = P A
P a +ρ g h a = P A
ρ g h a = P A - P a
h=5.15 m
The depth is 5.15 m.
Answer:
-30m/s
Explanation:
Given:
Initial velocity of object = 200 feet/second
Final velocity of object = 50 feet/second
Time of travel = 5 seconds
To calculate acceleration of the object we will find the rate of change of velocity with respect to time.
So, acceleration "a" is given by:
where vf represents final velocity, vi represents initial velocity and is time of travel.
Plugging in values to evaluate acceleration.
The acceleration of the object is -30m/s