Asked by ash
a rock is thrown vertically upward from ground level at t=0. At t= 1.5s, it passes the top of a tall tower and 1.0 s later, it reached its maximum height. what is the height of the tower?
Answers
Answered by
Henry
Tr = 1.5 + 1 = 2.5 s. = Rise time or time to reach max ht.
V = Vo + g*Tr = 0
Vo -9.8*2.5 = 0
Vo = 24.5 m/s. = Initial velocity.
h max = Vo*t + 0.5g*t^2 =
Vo*2.5 - 4.9*2.5^2 = 2.5Vo -
V = Vo + g*Tr = 0
Vo -9.8*2.5 = 0
Vo = 24.5 m/s. = Initial velocity.
h max = Vo*t + 0.5g*t^2 =
Vo*2.5 - 4.9*2.5^2 = 2.5Vo -
Answered by
Henry
Tr = 1.5 + 1 = 2.5 s. = Rise time or time to reach max ht.
V = Vo + g*Tr = 0
Vo -9.8*2.5 = 0
Vo = 24.5 m/s. = Initial velocity.
h = Vo*t + 0.5g*t^2 =
24.5*1.5 - 4.9*1.5^2 = 25.7 m. = Ht. of
the tower.
V = Vo + g*Tr = 0
Vo -9.8*2.5 = 0
Vo = 24.5 m/s. = Initial velocity.
h = Vo*t + 0.5g*t^2 =
24.5*1.5 - 4.9*1.5^2 = 25.7 m. = Ht. of
the tower.
Answered by
Henry
Please disregard my 5:24 PM post.
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