Question

julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a height of 1.27m from the ground. After how much time does Andrew catch the ball?

Answer 1

0.92

Step-by-step explanation:

Answer 2

Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2
Time to height
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s (ANSWER)