Question
A BALL IS PROJECTED VERTICALLY UPWARD FROM THE TOP OF A TOWER 60MITRE HIGH WITH A VELOCITY OF 30MITRE PER SECONDS.WHAT IS THE MAXIMUM HEIGHT ABOVE THE GROUND LEVEL?HOW LONG DOES IT TAKE TO REACH THE GROUND?
Answers
a. h = ho + (Vf^2-Vo^2)/2g.
ho = 60 m.
Vf = 0.
Vo = 30 m/s.
g = -9.8 m/s^2.
h = ?
b. Vf = Vo + g*Tr.
Vf = 0.
Vo = 30 m/s.
g = -9.8 m/s^2.
Tr = Rise time.
Tr = ?
b. 0.5g*Tf^2 = h(Part a).
g = 9.8 m/s^2.
Tf = Fall time.
Tf = ?
Tr + Tf = Time to reach gnd.
ho = 60 m.
Vf = 0.
Vo = 30 m/s.
g = -9.8 m/s^2.
h = ?
b. Vf = Vo + g*Tr.
Vf = 0.
Vo = 30 m/s.
g = -9.8 m/s^2.
Tr = Rise time.
Tr = ?
b. 0.5g*Tf^2 = h(Part a).
g = 9.8 m/s^2.
Tf = Fall time.
Tf = ?
Tr + Tf = Time to reach gnd.
,105
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