The speed of a projectile when it reaches its maximum height

[Ab17]

Homework Statement

The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maxi- mum height. What is the initial projection angle of the projectile?

Homework Equations

Vf2 = vi2 + 2ax

The Attempt at a Solution

I know that the speed at the top only consist of an x component and that the speed at half the height would be sqrt(vx2+vy2) . But I am trying everything from two hours and not getting a solution. The answer is suppost to be 67.8 but I don't know how to get it someone help

 

[Mastermind01]

The max height of the projectile is given by [itex] \frac {u^2sin^2(\theta)}{2g} [/itex] where [itex] \theta [/itex] is the angle of projection. Half the height would be [itex] \frac {u^2sin^2(\theta)}{4g} [/itex] . Find the y-component of velocity at half the height . Find the speed by adding the x and y components. Equate it to twice the x-component. Solve for [itex] \theta [/itex] . You should get your answer.

 

[Ab17]

X comp : Vicos@t
Y comp : sqrt(visin@ -2gh/2)

 

[Ab17]

X comp : Vicos@ ***

 

[Ab17]

Vf = sqrt((vicos@)^2 + Visin@-gh)

 

[Ab17]

Are those the correct components?

 

[Mastermind01]

The x-component is correct ([itex] v_icos (\theta) [/itex]) the y-component is wrong. You should learn [itex] \LaTeX [/itex] else your equations are hard to read.

 

[Ab17]

Visin@ - gt ?

 

[Mastermind01]

You are just giving me equations, I don't think you are putting in much effort. For the y-component [itex] v_i = v_isin(\theta) [/itex] , acceleration [itex] a = -g [/itex] and height [itex] h = \frac{v_i^2sin^2(\theta)}{4g} [/itex] What equation of motion do you use now to get [itex] v_f [/itex] ?

 

[Ab17]

Vf2 = vi2 + 2ax

 

[Ab17]

So I will get vi2sin2@/2 for vfy and then i get vf which will be sqrt(vi2cos2@ + vi2sin2@) i equate this to 2

 

[Ab17]

2viCos@ which gives me Cos@2 = 1/3 which gives an angle of 54.74 while the answer is 67.8

 

[Mastermind01]

You have clearly made a calculation mistake , using the equation you'll get [itex] v_f^2 = \frac{v_i^2sin^2(\theta)}{2} [/itex] . The resulting equation will be [tex] 4v_i^2cos^2(\theta) = v_i^2cos^2(\theta) + \frac{v_i^2sin^2(\theta)}{2} [/tex] Solve for [itex] \theta [/itex] you'll get your answer.