Asked by ashley
A golf ball is hit with an initial velocity of 50 m/s at an angle of 45° above the horizontal. How far will the ball travel horizontally before it hits the ground?
Answers
Answered by
Henry
Vo = (50m/s,45deg.).
Xo = hor. = 50cos45 = 35.36m/s.
Yo = ver. = 50sin45 = 35.36m/s.
t(up) = (Vf - Yo) / g,
t(up) = (0 - 35.36) / -9.8 = 3.61s.
t(dn) = t(up) = 3.61s.
T = t(up) + t(dn)=3.61 + 3.61=7.22s. =
Time in flight.
Dh = Xo * T = 35.36 * 7.22 = 255m. =
Hor. distance.
Alternate Method:
Dh = Vo^2 * sin2A / g.
A = 45 deg.
Vo = 50m/s.
Xo = hor. = 50cos45 = 35.36m/s.
Yo = ver. = 50sin45 = 35.36m/s.
t(up) = (Vf - Yo) / g,
t(up) = (0 - 35.36) / -9.8 = 3.61s.
t(dn) = t(up) = 3.61s.
T = t(up) + t(dn)=3.61 + 3.61=7.22s. =
Time in flight.
Dh = Xo * T = 35.36 * 7.22 = 255m. =
Hor. distance.
Alternate Method:
Dh = Vo^2 * sin2A / g.
A = 45 deg.
Vo = 50m/s.
Answered by
Anonymous
hyt
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.