A flying squirrel has a mass of 0. 765 kg. He jumps off the tree and hits the ground with 125 joules of energy. Determine how hi
1 answer:
Answer:
Height, h = 16.67 m
Explanation:
We have,
Mass of a squirrel is 0.765 kg.
He jumps off the tree and hits the ground with 125 joules of energy.
It is required to find the height up on the tree the squirrel was when it jumped.
The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :
h is height up on the tree the squirrel was when it jumped
So, the squirrel will go to a height of 16.67 m.
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The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>
The motion of the boulder is a uniformly accelerated motion, with constant acceleration
a = g = -9.8
downward (acceleration due to gravity).
By using Suvat equation:
v = u + at
where: v is the velocity at time t
u = 40.0 m/s is the initial velocity
a = g = -9.8 is the acceleration
To find the time t at which the velocity is v = 20.7 m/s
Therefore,
The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.
The complete question is:
A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?
To learn more about uniformly accelerated motion refer to:
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