Asked by Anonymous
Standing on the roof of a 62 m tall building, you throw a ball straight up with an initial speed of 14.5 m/s. If the ball misses the building on the way down, how long will it take from you threw the ball until it lands on the ground below? Give your answer in seconds and round the answer to three significant figures
Answers
Answered by
Henry
V^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g = -(14.5^2)/-19.6 = 10.73 m.
Above the roof.
V = Vo + g*Tr = 0.
Tr = -Vo/g = -14.5/-9.8 = 1.48 s. = Rise
time.
0.5g*Tf^2 = 62+10.73.
4.9*Tf^2 = 72.73.
Tf^2 = 14.8.
Tf = 3.85 s. = Fall time.
Tr+Tf = 1.48 + 3.85 = 5.33 s. = Time in
flight.
h = -Vo^2/2g = -(14.5^2)/-19.6 = 10.73 m.
Above the roof.
V = Vo + g*Tr = 0.
Tr = -Vo/g = -14.5/-9.8 = 1.48 s. = Rise
time.
0.5g*Tf^2 = 62+10.73.
4.9*Tf^2 = 72.73.
Tf^2 = 14.8.
Tf = 3.85 s. = Fall time.
Tr+Tf = 1.48 + 3.85 = 5.33 s. = Time in
flight.
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