Asked by Sean
A soccer ball is kicked with resultant velocity of 22.6 m/s at an angle of 60 degrees. What is the time to Apex, what is the range of the kick?
r = 22.6
angle = 60 degrees
1)Vxi=sin(60)(22.6) = 19.57
Vyi=cos(60)(22.6) = 11.3
2)T= Vyf-Vyi/A
0-11.3/-9.81 = 1.1 sec time to apex
Tup=Tdown (1.1+1.1) = 2.2 = total time
3)Range = Hxi(tt)
19.57(2.2) = 43.05 = range
Not sure if I went through this problem correctly. Please let me know if I made a mistake. Thanks!
r = 22.6
angle = 60 degrees
1)Vxi=sin(60)(22.6) = 19.57
Vyi=cos(60)(22.6) = 11.3
2)T= Vyf-Vyi/A
0-11.3/-9.81 = 1.1 sec time to apex
Tup=Tdown (1.1+1.1) = 2.2 = total time
3)Range = Hxi(tt)
19.57(2.2) = 43.05 = range
Not sure if I went through this problem correctly. Please let me know if I made a mistake. Thanks!
Answers
Answered by
Henry
Vo = 22.6 m/s[60o].
Xo = 22.6*Cos60 = 11.3 m/s.
Yo = 22.6*sin60 = 19.6 m/s.
Y = Yo + g*Tr = 0 @ Apex.
19.6 - 9.8*Tr = 0.
Tr = 2.0 s. = Rise time or time to reach
apex.
Tf = Tr = 2.0 s. = Fall time.
Range = Xo*(Tr+Tf) = 11.3m/s * 4s. =
45.2 m.
Note: Your procedure is correct, but you
made a mistake in step #1 which threw all of your answer off.
Xo = 22.6*Cos60 = 11.3 m/s.
Yo = 22.6*sin60 = 19.6 m/s.
Y = Yo + g*Tr = 0 @ Apex.
19.6 - 9.8*Tr = 0.
Tr = 2.0 s. = Rise time or time to reach
apex.
Tf = Tr = 2.0 s. = Fall time.
Range = Xo*(Tr+Tf) = 11.3m/s * 4s. =
45.2 m.
Note: Your procedure is correct, but you
made a mistake in step #1 which threw all of your answer off.
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