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A baseball is thrown up in the air from a height of 3 feet with an initial velocity of 23 feet per second. When does the baseba...Asked by Anonymous
A baseball is thrown up in the air from a height of 3 feet with an initial velocity of 23 feet per second. When does the baseball hit the ground?
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Answered by
Reiny
height = -16t^2 + 23t + 3
you want height = 0
so solve the quadratic
-16t^2 + 23t + 3 = 0
or
16t^2 - 23t - 3 = 0
Hint: it does not factor, so use the quadratic formula, reject the negative answer
you want height = 0
so solve the quadratic
-16t^2 + 23t + 3 = 0
or
16t^2 - 23t - 3 = 0
Hint: it does not factor, so use the quadratic formula, reject the negative answer
Answered by
henry2,
V = Vo + g*Tr = 0.
23 + (-32)Tr = 0,
Tr = 0.72 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2 = 3 + 23*0.72 + (-16)*0.72^2 = 11.3 Ft. above gnd.
h = 0.5g*Tf^2 = 11.3.
16Tf^2 = 11.3,
Tf = 0.84 s. = Fall time.
Tr + Tf = 0.72 + 0.84 = --------s. = Time to reach gnd.
23 + (-32)Tr = 0,
Tr = 0.72 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2 = 3 + 23*0.72 + (-16)*0.72^2 = 11.3 Ft. above gnd.
h = 0.5g*Tf^2 = 11.3.
16Tf^2 = 11.3,
Tf = 0.84 s. = Fall time.
Tr + Tf = 0.72 + 0.84 = --------s. = Time to reach gnd.
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