Asked by Arleen
a ball is thrown straight up with a speed of 8.0 m/s from a third-floor window located 13.0 m above ground. Calculate the maximum height, measured from the ground, of the ball; calculate the ball's velocity when it reaches the ground, and calculate the total time it takes the ball to reach the ground.
Answers
Answered by
Henry
a. ho + (V-Vo^2)/2g=13 + (0-(8^2)/-19.6 = 16.3 m. Above gnd.
b. V^2 = Vo^2 + 2g*h
Vo = 0 m/s, 2g = 19.6 m/s^2, h = 16.3 m.
Calculate V.
c. V = Vo + g*Tr = 0.
Tr = -Vo/g = -8/-9.8 = 0.82 s=Rise time.
h = 0.5g*t^2 = 16.3 m.
4.9t^2 = 16.3
t^2 = 3.33
Tf = 1.82 s. = Fall time.
Tr+Tf = 0.82 + 1.82 = 2.64 s. = Total
time to reach gnd.
b. V^2 = Vo^2 + 2g*h
Vo = 0 m/s, 2g = 19.6 m/s^2, h = 16.3 m.
Calculate V.
c. V = Vo + g*Tr = 0.
Tr = -Vo/g = -8/-9.8 = 0.82 s=Rise time.
h = 0.5g*t^2 = 16.3 m.
4.9t^2 = 16.3
t^2 = 3.33
Tf = 1.82 s. = Fall time.
Tr+Tf = 0.82 + 1.82 = 2.64 s. = Total
time to reach gnd.
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