Asked by Nami
A ball is thrown toward a cliff of height h with a speed of 27m/s and an angle of 60 degrees above horizontal. It lands on the edge of the cliff 3.0s later.
What is the height of the cliff?
What is the maximum height of the ball?
What is the ball's impact speed?
Please provide the Formulas and steps!
Thank You!
What is the height of the cliff?
What is the maximum height of the ball?
What is the ball's impact speed?
Please provide the Formulas and steps!
Thank You!
Answers
Answered by
Henry
Vo = 27m/s @ 60o
Xo = 27*cos60 = 13.5 m/s.
Yo = 27*sin60 = 23.4 m/s.
Tr = (Y-Yo)/g = (0-23.4)/-9.8 = 2.39 s.=
Rise time.
hmax = Yo*t + 0.5g*t^2.
hmax = 23.4*2.39 - 4.9*(2.39)^2=27.84 m.
Tr + Tf = 3.0 s.
2.39 + Tf = 3
Tf = 3 - 2.39 = 0.61 s. = Fall time.
h = hmax - 0.5g*Tf^2.
h = 27.84 - 4.9*(0.61)^2 = 23.3 = Ht. of cliff.
V = Vo + gt = 0 + 9.8*0.61 = 5.98 m/s.=
Impact velocity of ball.
Xo = 27*cos60 = 13.5 m/s.
Yo = 27*sin60 = 23.4 m/s.
Tr = (Y-Yo)/g = (0-23.4)/-9.8 = 2.39 s.=
Rise time.
hmax = Yo*t + 0.5g*t^2.
hmax = 23.4*2.39 - 4.9*(2.39)^2=27.84 m.
Tr + Tf = 3.0 s.
2.39 + Tf = 3
Tf = 3 - 2.39 = 0.61 s. = Fall time.
h = hmax - 0.5g*Tf^2.
h = 27.84 - 4.9*(0.61)^2 = 23.3 = Ht. of cliff.
V = Vo + gt = 0 + 9.8*0.61 = 5.98 m/s.=
Impact velocity of ball.
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