Asked by Hanna
a stone is thrown vertically upwards from the ground to reach a maximum height of 90m .how long will it take to reach a height of 75m
Answers
Answered by
Damon
well, it gets to 75 twice,once on the way up and once on the way down.
first find initial speed up, Vi
(1/2) m Vi*2= m g (90)
Vi = sqrt (2*9.81*90) = 42.0 m/s
then
v = 42 - .81 t
h = 0 + 42 t - 4.9 t^2 = 75
4.9 t^2 - 42 t + 75 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
first find initial speed up, Vi
(1/2) m Vi*2= m g (90)
Vi = sqrt (2*9.81*90) = 42.0 m/s
then
v = 42 - .81 t
h = 0 + 42 t - 4.9 t^2 = 75
4.9 t^2 - 42 t + 75 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
Answered by
henry2,
V^2 = Vo^2 + 2g*h = 0.
Vo^2-19.6*90 = 0
Vo = 42 m/s.
V^2 = Vo^2+2g*h = 42^2+(-19.6)75 = 294
V = 17.1 m/s. at 75 m.
V = Vo+g*T = 17.1
42-9.8T = 17.1
T = 2.5 s. to reach 75 m.
Vo^2-19.6*90 = 0
Vo = 42 m/s.
V^2 = Vo^2+2g*h = 42^2+(-19.6)75 = 294
V = 17.1 m/s. at 75 m.
V = Vo+g*T = 17.1
42-9.8T = 17.1
T = 2.5 s. to reach 75 m.
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