An egg is thrown nearly vertically upward from a point near the cornice of a tall building. It just misses the cornice on the way down and passes a point 50.0m below its starting point 6.00 seconds after it leaves the thrower's hand. Air resistance may be ignored.
a.) what is the initial speed of the egg? (answer: 21.2 m/s)
b.)How high does it rise above its starting point?(answer: 22.6m)
im stuck with these two questions and i got 58m/s in the 1st part ...im reallyyy stuck
You need the equation of motion which if we takeOriginally Posted by ^_^Engineer_Adam^_^
positive
upward and the origin as the point from which it is threown we have:
so:
and :
We are told thatat
so:
so:
It reaches its maximum height whenwhich is when:
RonL
Hello, ^_^Engineer_Adam^_^!
An egg is thrown nearly vertically upward from a point near the cornice of a tall building.
It just misses the cornice on the way down and passes a point 50.0m below its starting point
6.00 seconds after it leaves the thrower's hand.
a) What is the initial speed of the egg? (answer: 21.2 m/s) ?
b) How high does it rise above its starting point? (answer: 22.6m)
Are you familiar with the free-fall formula? .
. . where= initial height,
= initial velocity,
. . and= height of object at time
.
Let launch point be at.
. . Then we have: .
(a) We are told thatwhen
.
. . So we have: .
Therefore, the initial speed is: .
(b) To maximize height, equate the derivative to zero.
We have: .
Then: .seconds
Therefore: .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
If you aren't allowed to use Calculus, we can still solve it.
The graph of: .is a down-opening parabola.
. . Its maximum height occurs at its vertex.
The vertex is given by: .
We have: .
. . Hence: .seconds . . . see?
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