SOLUTION: A stone is thrown upward with an initial velocity of 15.24 m/s. What will its maximum height be? When will it strike the ground? Where will it be in 1.30 s?
Thankyou in advance
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Question 1123908: A stone is thrown upward with an initial velocity of 15.24 m/s. What will its maximum height be? When will it strike the ground? Where will it be in 1.30 s?
Thankyou in advance!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A stone is thrown upward with an initial velocity of 15.24 m/s.
What will its maximum height be?
:
Using -4.9 m/sec as the force of gravity
h = -4.9x^2 + 15.24x
Max height occurs at the axis of symmetry x=-b/(2a), a = -4.9, b=15.24
x =
x = 1.555 sec it will be max height
Find the actual height
h = -4.9(1.555^2) + 15.24(1.555)
h = -11,848 + 23.698
h = +11.7 meter is max height
:
When will it strike the ground?
It strikes the ground when x=0
-4.9x^2 + 15.24x = 0
Factor out x
x(-4.9x + 15.24)
two solutions
x = 0 initial position
and
-4.9x = -15.24
x = -15.24/-4.9
x = 3.11 sec to hit the ground
:
Where will it be in 1.30 s?
x=1.3
h = -4.9(1.3^2) + 15.24(1.3)
I'll let you calculate this
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