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.
ANSWER :
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
V_{y}^{2} =V_{oy}^{2}+2a_{y}*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*(V_{y}-V_{oy})/a =2*(0-10sin(o))/(-9.8)
t=-20Sin(o)/9.8
the range of projectile is
dX =V_{ox}t =(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) =74.5^{o}
ANSWER :
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).
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> 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