Asked by Aquile
A ball is thrown upwards at 15m/s from a window 20 metres above the ground. Find: a)How long the ball the takes to reach the level of the window
b)How much longer it takes to reach the ground?
c)What speed it takes to reach the ground?
b)How much longer it takes to reach the ground?
c)What speed it takes to reach the ground?
Answers
Answered by
Henry
a. = V = Vo + g*t
Tr = (V-Vo)/g = (0-15)/-9.8 = 1.53 s. =
Rise time.
Tf = Tr = 1.53 s. = Fall time.
T = Tr + Tf = 1.53 + 1.53 = 3.06 s. =
Time to return to window level.
b. H = Vo*t + 0.5g*t^2 = 20 m.
15*t + 4.9t^2 = 20
4.9t^2 + 15t - 20 = 0
Use Quadratic Formula.
T = 1.0 s. To reach Gnd.
c. V^2 = Vo^2 + 2g*h.
V^2 = 15 + 19.6*20 = 392
V = 19.8 m/s.
Tr = (V-Vo)/g = (0-15)/-9.8 = 1.53 s. =
Rise time.
Tf = Tr = 1.53 s. = Fall time.
T = Tr + Tf = 1.53 + 1.53 = 3.06 s. =
Time to return to window level.
b. H = Vo*t + 0.5g*t^2 = 20 m.
15*t + 4.9t^2 = 20
4.9t^2 + 15t - 20 = 0
Use Quadratic Formula.
T = 1.0 s. To reach Gnd.
c. V^2 = Vo^2 + 2g*h.
V^2 = 15 + 19.6*20 = 392
V = 19.8 m/s.
Answered by
Henry
Correction:
c. V^2 = (15)^2 + 19.6*20 = 617
V = 24.8 m/s.
c. V^2 = (15)^2 + 19.6*20 = 617
V = 24.8 m/s.
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