What Angle of Projection Makes Maximum Height Equal to Range?

  • Thread starter Thread starter kuakkgom
  • Start date Start date
  • Tags Tags
    Projectile
AI Thread Summary
To determine the angle of projection where maximum height equals range for a projectile launched at 53 m/s, the key formulas are R = Vo^2 * sin(2x) / g and H = Voy^2 / (2g). Setting maximum height (H) equal to range (R) leads to the equation Voy^2 = 2 * Vo^2 * sin(2x). The challenge lies in isolating the angle x, as it is necessary to break down the initial velocity into its components. The discussion highlights the difficulty in solving for x without additional information or steps.
kuakkgom
Messages
1
Reaction score
0
This problem has been giving me headaches for the past two weeks. Here it is, as well as how far I have gotten on it:

A projectile is fired with an initial velocity of 53m/s. Find the angle of projection such that the maximum height is equal to it's range.

So that basically boils down to:

H = R
Vo = 53m/s
ay = -9.81m/s^2
t = unknown
x (angle) = find

According to the nifty sheet the teacher tossed out the applicable formulas would be:

R=Vo^2*sin(2x)/g (since Ho and Hf will be the same)
H=Voy^2/2g.

Making H=R gets the formula Voy^2=2*Vo^2*sin(2x) ...which doesn't help me since x is what needs to be found, and without x one can't break Vo into component form.

Do I have the right idea or have I flubbed it already?

Thanks,

-K
 
Physics news on Phys.org
Voy=vo*sinx
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

How Do You Calculate Maximum Height and Range of a Projectile?
Replies
1
Views
1K
What is the Launch Angle of a Projectile at Half Its Maximum Height?
Replies
3
Views
2K
How Do You Solve a Projectile Motion Problem?
Replies
3
Views
4K
Needing some help about time of flight of a launched ball/horiz range
Replies
6
Views
3K
Find the height of a lamp based on the velocities of projectiles
Replies
40
Views
2K
Find Initial Velocity Given Height, Range, and Initial Angle
Replies
4
Views
2K
Projectile Motion - vertical displacement at specific time.
Replies
3
Views
2K
Projectile Motion: Finding the Maximum Height of a Football
Replies
5
Views
6K
Maximum range on ground for an elevated cannon
Replies
21
Views
4K
What Is the Angle Between Velocity Vector and Trajectory at Maximum Height?
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top