Asked by onyebuchi ogor
A ball is thrown with an initial velocity of 20mls at an angle of 30 to the horizontal calculate time of flight maximum height the horizontal range of the ball
Answers
Answered by
Henry
Vo = 20m/s[30o].
Xo = 20*Cos30 = 17.32 m/s.
Yo = 20*sin30 = 10 m/s.
a. Y = Yo + g*Tr.
0 = 10 -9.8Tr, Tr = 1.02 s. = Rise time.
Tf = Tr = 1.02 s.
Tr+Tf = 1.02 + 1.02 = 2.04 s. = Time in flight.
b. h = Yo*Tr + 0.5g*Tr^2.
h = 10*1.02 - 4.9*1.02^2 =
c. Range = Xo*(Tr+Tf)= 17.32*2.04 =
Xo = 20*Cos30 = 17.32 m/s.
Yo = 20*sin30 = 10 m/s.
a. Y = Yo + g*Tr.
0 = 10 -9.8Tr, Tr = 1.02 s. = Rise time.
Tf = Tr = 1.02 s.
Tr+Tf = 1.02 + 1.02 = 2.04 s. = Time in flight.
b. h = Yo*Tr + 0.5g*Tr^2.
h = 10*1.02 - 4.9*1.02^2 =
c. Range = Xo*(Tr+Tf)= 17.32*2.04 =
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