Maximum height of projectile motion

[Dru2k]

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]

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]

You could also use the range equation to solve for initial velocity. Then use that to find max height.

 

[Dru2k]

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?

 

[Dru2k]

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]

Your calculations are correct!

-- AI

 

[shimoda1]

I found this video helpful! check it out

 

[shimoda1]

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