View Full Version : Maximum height of projectile motion
I'm having major problems figuring this one out.
The question is from the book, yet their are no sample problems and a handful of given formulas. I've played around with the formulas with substitution, etc and I've end up with the wrong answer.
Given: Angle of projectile fired and final displacement along X axis.
angle = 25 degrees
final displacement of x = 301.5 m
accleration = -9.81m/s^2
Find: Maximum height of Y.
any ideas how I can manipulate the standard formulas of projectile motion to solve for this problem?
Chi Meson
Oct3-04, 07:40 PM
horizontal displacement = v(cos theta) t
vertical displacement = 0 = v(sin theta)t + 1/2 (-9.8) t^2
re-arrange each formula, solving for t, and set them equal to each other.
THen solve for v
then solve for max height.
Parth Dave
Oct3-04, 07:40 PM
You could also use the range equation to solve for initial velocity. Then use that to find max height.
can you help me find my mistake. i worked it out to the best of my ability.
rearranging displacement of x
t = 301.5m / v cos 25
rearranging displacement of y
t = -2 ( v sin25) / -9.81
than using t = t
-2 ( v sin 25) ( v cos 25) = 301.5 (-9.81) -->
V^2 = sqroot [ (301.5(-9.81)) / -2 sin25 cos 25
V = 62.1 m/s
T = 301.5m / 62.1m/s cos 25 = 5.36s
maximum height with no air resistance would be at half of time
displacement of y = v sin 25 ( t ) + (1/2) (-9.81) (t^2)
y = 62.1 sin 25 (2.68) + 1/2 (-9.81) (2.68^2)
maximum y = 70.3 - 35.2 = 35.1
answer in book says it is 70.3m
what did i do wrong?
maybe this will help
Exact question from book: A golfer can hit a golf ball a horizontal distance of over 300 m on a good drive. What maximum height will a 301.5m drive reach if it is launched at an angle of 25.0 degrees to the ground? (Hint: At the top of its flight, the ball's vertical velocity component will be zero)
TenaliRaman
Oct4-04, 06:26 AM
Your calculations are correct!
-- AI
shimoda1
Sep29-11, 02:50 PM
http://www.youtube.com/watch?v=q1lEJVhS_5g
I found this video helpful! check it out
shimoda1
Sep29-11, 02:50 PM
http://www.youtube.com/watch?v=q1lEJVhS_5g
I found this video helpful! check it out
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