Asked by Doug
a golf ball is observed to hit the ground 800 ft. away after a flight of 5 seconds. Assume that the ground is level and neglect friction of the air. Find the angle of elevation above the ground of the initial motion of the ball, find the initial velocity of the ball, find its maximum height above t he ground v = v + gt v = 160 ft/sec ?
or 800 = v(5) + 1/2(32)(25) = 240 ft/sec
or 800 = v(5) + 1/2(32)(25) = 240 ft/sec
Answers
Answered by
Henry
D = Xo * t = 800 Ft.
Xo * 5 = 800
Xo = 160 Ft/s.
Tr + Tf = 5 s. Tr = Tf.
Tr+Tr = 5.
Tr = 2.5 s. = Rise time.
Y = Yo + g*Tr = 0.
Yo = -g*Tr = -(-32)*2.5 = 80 Ft/s.
1. Tan A = Yo/Xo = 80/160 = 0.50.
A = 26.6o.
2. Vo = Xo + Yoi = 160 + 80i = 179Ft/s[26.6o].
3. h = Yo*Tr + 0.5g*Tr^2.
g = (-)32 Ft/s^2.
Solve for h.
Xo * 5 = 800
Xo = 160 Ft/s.
Tr + Tf = 5 s. Tr = Tf.
Tr+Tr = 5.
Tr = 2.5 s. = Rise time.
Y = Yo + g*Tr = 0.
Yo = -g*Tr = -(-32)*2.5 = 80 Ft/s.
1. Tan A = Yo/Xo = 80/160 = 0.50.
A = 26.6o.
2. Vo = Xo + Yoi = 160 + 80i = 179Ft/s[26.6o].
3. h = Yo*Tr + 0.5g*Tr^2.
g = (-)32 Ft/s^2.
Solve for h.
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