Answers and Solutions

Answer

Answer:

Let the projectile  be launched  with an initial velocity of V0 at an angle of  x degrees to the horizontal  ground.

The vertical component of the velocity U0 = V0 sin x

Horizontal component of velocity =  H  =     V0  cos x

As the projectile  goes up ,its vertical  component of velocity  diminishes but the horizontal component of the velocity  remains constant.

 The vertical component of the velocity U at any height h,  is given by the formula 

(U)^2  - (U0)^2  =  -2gh. where g is acceleration due to gravity. Let h1 be the maximum height reached by the projectile. At maximum height , the vertical component of the velocity  

U1 becomes 0 ,so we get  0- (U0)^2 =  -2gh1. Therefore     h1 = (U0)^2 / (2g)  

Now h2 =  half maximum height =  (h1)/2  =   (U0)^2 / (4g)

Therefore if U2   = vertical component of the velocity at height h2, then

(U2)^2  - (U0)^2  =  -2gh2   or  (U2)^2  = (U0)^2  - 2gh2  =   (U0)^2 - 2g*(U0)^2/ 4g = 

(U0)^2 -  (U0)^2/2  =  (U0)^2/2 .

Thus (U1)^2 =  0 and

(U2)^2 =  (U0)^2/2  

The horizontal component of  the velocity  H = V0  cos x   does not undergo any deceleration and so remains constant throughout.

Hence total Velocity of the projectile   at maximum height of h1  , V1 =  √{0+(H)^2} = H

So   total velocity at maximum height h1 ,    V1 =  V0  cos x

Total Velocity of the projectile   at half maximum height h2,       V2 = √{(U2)^2 + H^2}=

= √ { (U0)^2/2 + H^2}  = √ {( V0 sin x)^2/2 +( V0  cos x )^2}    =  V0 √{ ( sin x)^2/2 +(cos x)^2}=

V0 √[{ 1 - (cos x)^2+2(cos x)^2}/2]   =  V0 √[{ 1+(cos x)^2}/2] .

So total velocity of the projectile  at half maximum height -

V2 =    V0 √[{ 1+(cos x)^2}/2]

Therefore  V1/V2   =    V0  cos x / V0 √[{ 1+(cos x)^2}/2]   = √(2/5 ) 

Or  (V1/V2)^2 =    2(cos x)^2/{ 1+(cos x)^2  =   2/5. 

Simplifying 5(cos x)^2  =  1+(cos x)^2  or 4( cos x)^2  = 1 .So (cos x)^2 =  1/ 4 

Hence Cos x  =  1/ 2 and  so  x = 60° 

Therefore the angle of projection = 60°