# Help me with the question?

##### 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 #