Asked by Mariam
A pumpkin is thrown upwards at a speed of 27.5 m/s and from a height of 2 m above the ground. How long before it hits the ground and smashes?
Not sure where to begin with this.
Not sure where to begin with this.
Answers
Answered by
Anonymous
h = Hi + Vi t - 4.9 t^2
0 = 2 + 27.5 t -4.9 t^2
or
4.9 t^2 -27.5 t - 2 = 0
solve quadratic for t, use the positive answer (any negative root was on the way up before you threw it :)
0 = 2 + 27.5 t -4.9 t^2
or
4.9 t^2 -27.5 t - 2 = 0
solve quadratic for t, use the positive answer (any negative root was on the way up before you threw it :)
Answered by
Henry
V^2 = Vo^2 + 2g*h.
0 = 27.5^2 - 19.8h
h = 38.6m. above launching point.
ho + h = 2 + 38.6 = 40.6m above gnd.
V = Vo + g*Tr.
0 = 27.5 - 9.8Tr
Tr = 2.81s. = Rise time.
0.5g*Tf^2 = 40.6.
Tf(fall time) = ?.
Tr+Tf = Time to reach gnd.
0 = 27.5^2 - 19.8h
h = 38.6m. above launching point.
ho + h = 2 + 38.6 = 40.6m above gnd.
V = Vo + g*Tr.
0 = 27.5 - 9.8Tr
Tr = 2.81s. = Rise time.
0.5g*Tf^2 = 40.6.
Tf(fall time) = ?.
Tr+Tf = Time to reach gnd.
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