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1 Answer
# alpha = 60^o #
Explanation:
For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:
#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #
Let us define the following variables:
# { (U, "= Initial Speed", ms^-1), (alpha, "= Angle of projection to the horizontal", ""^o), (h, "= Maximum Height", m), (U_h, "= Horizontal Speed (constant)", ms^-1), (V_("top"), "= Overall speed at highest point", ms^-1), (V_("half"), "= Overall speed at half of maximum height", ms^-1) :} #
Horizontal Motion
The projectile will move under constant speed (NB we can still use "suvat" equation with
# U_h = Ucosalpha#
Vertical Motion
The projectile travels under constant acceleration due to gravity.
When the projectile reaches its maximum height Its vertical speed will momentarily be zero, so we can calculate the maximum height:
# { (s=,h,m),(u=,Usin alpha,ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,"Not Required",s) :} #
Applying
# 0^2 = (Usinalpha)^2 + (2)(-g)(h) #
# :. 2gh = U^2sin^2alpha #
# :. h = (U^2sin^2alpha)/(2g) #
The overall speed at this maximum height is just the horizontal component, thus:
# V_("top") = Ucos alpha #
Now, let us calculate the vertical speed at halfway of the maximum height, ie
# { (s=,(U^2sin^2alpha)/(4g),m),(u=,Usin alpha,ms^-1),(v=,v,ms^-1),(a=,-g,ms^-2),(t=,"Not Required",s) :} #
Applying
# v^2 = (Usinalpha)^2+2(-g)((U^2sin^2alpha)/(4g)) #
# \ \ \ = U^2sin^2alpha - (U^2sin^2alpha)/(2) #
# \ \ \ = (U^2sin^2alpha)/(2) #
So, the overall speed at this half way height is given by:
# V_("half")^2 = v^2 + U_h^2 #
# \ \ \ \ \ \ = (U^2sin^2alpha)/(2) + U^2cos^2alpha#
# :. V_("half") = sqrt( (U^2sin^2alpha)/(2) + U^2cos^2alpha )#
The next step is to use the fact that the speed at the maximum height is
# => V_("top") = sqrt(2/5) xx V_("half") #
# :. V_("top")^2 = 2/5 xx V_("half")^2 #
And using the values determined by the above calculation:
# (Ucos alpha)^2 = 2/5 {(U^2sin^2alpha)/(2) + U^2cos^2alpha } #
# :. (5U^2cos^2 alpha)/2 = (U^2sin^2alpha)/(2) + U^2cos^2alpha #
# :. (5cos^2 alpha)/2 = (sin^2alpha)/(2) + cos^2alpha #
# :. 5cos^2 alpha = sin^2alpha + 2cos^2alpha #
# :. 3cos^2 alpha = sin^2alpha #
# :. sin^2alpha/cos^2 alpha = 3#
# :. tan^2alpha = 3#
# :. tanalpha = +-sqrt(3)#
However, as
# tanalpha = sqrt(3)#
Which then gives us:
# alpha = 60^o #