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

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)?

#### Solution

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°

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Chapter 4: Motion In a Plane - Exercises [Page 27]

#### APPEARS IN

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