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
(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) What will be time of flight?
(c) What is the distance (horizontal range) from the point of projection at which the ball will land ?
(d) Find q at which he should throw the ball that would maximise the horizontal range as found in (iii).
(e) How does q for maximum range change if u >vo , u = vo , u < vo?
(f) How does q in (v) compare with that for u = 0 (i.e.450) ?
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
(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.
(b) What will be time of flight?
(c) What is the distance (horizontal range) from the point of projection at which the ball will land ?
(d) Find q at which he should throw the ball that would maximise the horizontal range as found in (iii).
(e) How does q for maximum range change if u >vo , u = vo , u < vo?
(f) How does q in (v) compare with that for u = 0 (i.e.450) ?
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1 Answer
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This is a Long Answer Type Question as classified in NCERT Exemplar
Explanation- a) for x direction ux= u+vocos
uy=velocity in y direction= v0sin
now tan
b) let t be the time flight y =0 uy=vosin
y= uyt+1/2 ayt2
0= vosin +
So T =
c) horizontal range R, = (u+vocos T= (u+vocos )
d) for range to be maximum dR/d
4vocos2
So cos =
e) cos =
so
f) if u=0 0
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Please find the solution below:
after 10 kicks,
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
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
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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