4.32

(a) Show that for a projectile the angle between the velocity and the x-axis as a function

of time is given by

(b) Shows that the projection angle _0 for a projectile launched from the origin is

given by

where the symbols have their usual meaning.

0 44 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    4.32

    (a) Let θt=tan-1?voy-gtvox be the angle at which the projectile is fired w.r.t. the x-axis, since θ0=tan-1?4hmR depends on t

    Therefore tan θ since (Vy=Voy -gt and Vx = Vox)

    (b) Since θ = θt=VxVy=Voy-gtVox sin2 θt=tan-1?(Voy-gtVox) /2g…….(1)

    R = hmax sin2 u2 /g …….(2)

    Dividing (1) by (2)

    θ /R) = [ u2 sin2 θ /2g]/[ hmax sin2 u2 /g] = θ / 4

Similar Questions for you

A
alok kumar singh

Please find the solution below:

 

 

 

V
Vishal Baghel

  a = v 0 2 R

after 10 kicks, v = v 0 + 1 0 v 0 1 0 0 = 1 . 1 v 0

R 1 = 1 . 2 1 R

 

V
Vishal Baghel

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

V
Vishal Baghel

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

V
Vishal Baghel

a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r

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