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# A projectile is fired with an initial speed of 10 m/s. Find theangle of projection...

A projectile is fired with an initial speed of 10 m/s. Find the angle of projection such that the maximum height of the projectile is equal to 0.9 times its horizontal range.

Initial speed = 10 m/s

Let the angle of projection be with the horizontal.

So, vertical component of initial speed = 10 sin

So,

Total K.E. initially = 1/2 m (10)^2  (independent of ?

K.E. in the vertical direction = 1/2 m (10 sin ?)^2

K.E. in the horizontal direction = 1/2 m (10 cos ?)^2

At maximum height, K.E in vertical direction = 0  (since velocity in the vertical direction is 0)

but it gains P.E. by gaining the height h.

So,

Gain of P.E = Loss of K.E.

=> m g h max = 1/2  m (10 sin ?)^2

=> h max = (10 sin ? )^2  / 2g  =  (50/9.8) sin^2 ? = 5.102 sin^2 ?

Time taken for attaining h max =  h max / average speed  = 5.102 sin^2 ? / (10 sin ? + 0)/2 = 1.0204 sin ?

Distance travelled horizontally when height is  h max = 10 cos ? * 1.0204 sin ? = 10.204 sin ? cos ?

Projectile reaches maximum height and comes back down and hits the ground.

So,

Total horizontal distance travelled

= 2 * horizontal distance travelled when at h max

= 2 * 10.204 sin ? cos ?

= 20.408 sin ? cos ?

Given that :

h max = 0.9 total horizontal distance travelled

=> 5.102 sin^2 ? = 0.9 * 20.408 sin ? cos ?

=> sin ? / cos ? = 0.9 * 20.408 / 5.102

=> tan ? = 3.6

=> ? = 74.48 º  (with the horizontal)  (ANSWER).

The Maximum height reached by the projectile is given by

Vy2 =Voy2+2ay*dY

Since final Velocity is zero

0=(10Sin(o))2+2*(-9.8)*dY

dY =(10Sin(o))2/19.6

The time of flight the projectile is twice time needed to reach the maximum height

t=2*(Vy-Voy)/a =2*(0-10sin(o))/(-9.8)

t=-20Sin(o)/9.8

the range of projectile is

dX =Voxt =(10Cos(o))*20Sin(o)/9.8

given

dY =0.9dX

(10Sin(o))2/19.6 =0.9*(10Cos(o))*20Sin(o)/9.8

tan(o) =3.6

o=tan-1(3.6) =74.5o

Initial speed = 10 m/s

Let the angle of projection be with the horizontal.

So, vertical component of initial speed = 10 sin

So,

Total K.E. initially = 1/2 m (10)^2  (independent of ?

K.E. in the vertical direction = 1/2 m (10 sin ?)^2

K.E. in the horizontal direction = 1/2 m (10 cos ?)^2

At maximum height, K.E in vertical direction = 0  (since velocity in the vertical direction is 0)

but it gains P.E. by gaining the height h.

So,

Gain of P.E = Loss of K.E.

=> m g h max = 1/2  m (10 sin ?)^2

=> h max = (10 sin ? )^2  / 2g  =  (50/9.8) sin^2 ? = 5.102 sin^2 ?

Time taken for attaining h max =  h max / average speed  = 5.102 sin^2 ? / (10 sin ? - 0)/2 = 1.0204 sin ?

Distance travelled horizontally when height is  h max = 10 cos ? * 1.0204 sin ? = 10.204 sin ? cos ?

Projectile reaches maximum height and comes back down and hits the ground.

So,

Total horizontal distance travelled

= 2 * horizontal distance travelled when at h max

= 2 * 10.204 sin ? cos ?

= 20.408 sin ? cos ?

Given that :

h max = 0.9 total horizontal distance travelled

=> 5.102 sin^2 ? = 20.408 sin ? cos ?

=> sin ? / cos ? = 20.408 / 5.102

=> tan ? = 4

=> ? = 75.964 º  (with the horizontal)  (ANSWER).

> Please correct the last 4 lines as under :
=> 5.102 sin^2 𝛉 = 0.9 * 20.408 sin 𝛉 cos 𝛉
=> sin 𝛉 / cos 𝛉 = 0.9 * 20.408 / 5.102
=> tan 𝛉 = 3.6
=> 𝛉 = 74.48 º (with the horizontal) (ANSWER).

Tulsiram Garg Tue, Oct 26, 2021 8:29 AM

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