a₀). You know ...
-- the object is dropped from 5 meters
above the pavement;
-- it falls for 0.83 second.
a₁). Without being told, you assume ...
-- there is no air anyplace where the marshmallow travels,
so it free-falls, with no air resistance;
-- the event is happening on Earth,
where the acceleration of gravity is 9.81 m/s² .
b). You need to find how much LESS than 5 meters
the marshmallow falls in 0.83 second.
c). You can use whatever equations you like.
I'm going to use the equation for the distance an object falls in
' T ' seconds, in a place where the acceleration of gravity is ' G '.
d). To see how this all goes together for the solution, keep reading:
The distance that an object falls in ' T ' seconds
when it's dropped from rest is
(1/2 G) x (T²) .
On Earth, ' G ' is roughly 9.81 m/s², so in 0.83 seconds,
such an object would fall
(9.81 / 2) x (0.83)² = 3.38 meters .
It dropped from 5 meters above the pavement, but it
only fell 3.38 meters before something stopped it.
So it must have hit something that was
(5.00 - 3.38) = 1.62 meters
above the pavement. That's where the head of the unsuspecting
person was as he innocently walked by and got clobbered.
Work = force in the direction of the movement x distance = 27 N x 1.7 m
Work = 45.9 joules
Answer: option c.
B. Kinetic energy plus its potential energy.
1.49 MPH (Miles Per Hour)
Answer:
4.5 m/s
Explanation:
The rock must barely clear the shelf below, this means that the horizontal distance covered must be
while the vertical distance covered must be
The rock is thrown horizontally with velocity 
, so we can rewrite the horizontal distance as
where t is the time of flight. Re-arranging the equation,
(1)
The vertical distance covered instead is
where we omit the term 
since the initial vertical velocity is zero. From this equation,
(2)
Equating (1) and (2), we can solve the equation to find :
