Asked by sydney
A water balloon is thrown into the air and lands 30.0 meters away 1.8 seconds later.
a. What was the speed of the balloon thrown (on the diagonal)
b. At what angle was the balloon thrown with respect to the ground?
a. What was the speed of the balloon thrown (on the diagonal)
b. At what angle was the balloon thrown with respect to the ground?
Answers
Answered by
Henry
D = Xo * 1.8 = 30 m.
Xo = 30/1.8 = 16.67 m/s=Hor. component of initial velocity.
Y = Yo + g*Tr = 0 at max Ht.
Yo - 9.8*(1.8/2) = 0
Yo - 8.82 = 0
Yo = 8.82 m/s = Ver. component of initial velocity.
a. Vo^2=Xo^2 + Yo^2=16.67^2 + 8.82^2 =
355.7
Vo = 18.9 m/s = Initial velocity.
b. tan A = Yo/Xo = 8.82/16.67 = 0.52909
A = 27.9o
Xo = 30/1.8 = 16.67 m/s=Hor. component of initial velocity.
Y = Yo + g*Tr = 0 at max Ht.
Yo - 9.8*(1.8/2) = 0
Yo - 8.82 = 0
Yo = 8.82 m/s = Ver. component of initial velocity.
a. Vo^2=Xo^2 + Yo^2=16.67^2 + 8.82^2 =
355.7
Vo = 18.9 m/s = Initial velocity.
b. tan A = Yo/Xo = 8.82/16.67 = 0.52909
A = 27.9o
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