# 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?

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Chi Meson
Homework Helper
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.

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)

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