A cricket fielder can throw the cricket ball with a speed vo. - Physics

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Long Answer

A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle θ to the horizontal, find 

  1. the effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
  2. what will be time of flight?
  3. what is the distance (horizontal range) from the point of projection at which the ball will land?
  4. find θ at which he should throw the ball that would maximise the horizontal range as found in (iii).
  5. how does θ for maximum range change if u > vo, u = vo, u < vo?
  6. how does θ in (v) compare with that for u = 0 (i.e.45)?
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Solution

Consider the adjacent diagram,

a. Initial velocity in x-direction, `u_x = u + v_0  cos theta`

`u_y` = Initial velocity in y-direction

= `v_0  sin theta`

Where the angle of projection is θ.

Now, we can write

`tan theta = u_y/U_x = (u_0  sin  theta)/(u + u_0  cos theta)`

⇒ `θ = tan^-1 ((v_0  sin  theta)/(u + v_0  cos theta))`

b. Let T be the time of the flight.

As net displacement is zero over time period T.

 `y = 0, u_y = v_0 sin θ, a-y = - g, t = T`

We know that `y = u_yt + 1/2 a_yt^2`

⇒ 0 = `v_0  sin θ  T + 1/2 (- g) T^2`

⇒ `T[v_0  sin θ - g/2 T]` = 0

⇒ T = `0, (2v_g  sin θ)/g`

T = 01, which corresponds to point O.

Hence, T = `(2u_0 sin θ)/g`

c. Horizontal range, `R(u + v_0 cos θ)`

`T = (u + v_0 cos θ) (2v_0 sin θ)/g`

= `v_0/g [2u sin θ + v_0 sin 2θ]`

d. For horizontal range to be maximum , `(dR)/(dθ)` = 0

⇒ `v_0/g [2u cos θ + v_0 cos 2θ xx 2]` = 0

⇒ `2u cos θ + 2v_0 [2cos^2θ - 1]` = 0

⇒ `4v_0 cos^2θ + 2u cos θ - 2v_0` = 0

⇒ `2v_0 cos^2θ + u cos θ - v_0` = 0

⇒ `cos θ = (-u +- sqrt(u^2 + 8v_0^2))/(4v_0)`

⇒ `θ_"max" cos^-1 [(-u +- sqrt(u^2 + 8v_0^2))/(4v_0)]`

= `cos^-1 [(-u + sqrt(u^2 + 8v_0^2))/(4v_0)]`

e. If u = v0,

`cos θ = (-v_0 +- sqrt(v_0^2 + 8v_0^2))/(4v_0) = (-1 + 3)/4 = 1/2`

If `u < < v_0`, then `8v_0^2 + u^2 ≈ 8v_0^2`

`θ_"max" = cos^-1[(-u +- 2sqrt(2)v_0)/(4v_0)] = cos^-1[1/sqrt(2) - u/(4v_0)]`

If `u < < v_0`, then `θ_"max" = cos^-1 (1/sqrt(2)) = π/4`

If `u > u_0` and `u > > v_0`

`θ_"max" = cos^-1 [(-u +- u)/(4v_0)]` = 0

⇒ `θ_"max" = π/2`

f. If u = 0, `θ_"max" = cos^-1 [(0 +- sqrt(8v_0^2))/(4v_0)]`

= `cos^-1 (1/sqrt(2))`

= 45°

  Is there an error in this question or solution?
Chapter 4: Motion In a Plane - Exercises [Page 27]

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NCERT Exemplar Physics Class 11
Chapter 4 Motion In a Plane
Exercises | Q 4.35 | Page 27

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