A cricket fielder can throw the cricket ball with a speed v_o . 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 >v_o , u = v_o , u < v_o?

(f) How does q in (v) compare with that for u = 0 (i.e.45⁰) ?

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Vishal Baghel | Contributor-Level 10

7 months ago

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- a) for x direction u_x= u+v_ocos $θ$

u_y=velocity in y direction= v₀sin $θ$

now tan $θ = \frac{u_{y}}{u_{x}} = \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

$θ = {t a n}^{- 1} \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

b) let t be the time flight y =0 u_y=v_osin $θ, a_{y} = - g, t = T$

y= u_yt+1/2 a_yt²

0= v_osin $θ T$ + $\frac{1}{2} (- g T^{2})$

So T = $\frac{2 u_{o} s i n θ}{g}$

c) horizontal range R, = (u+v_ocos $θ$ T= (u+v_ocos $θ$ ) $\frac{2 u_{o} s i n θ}{g}$

d) for range

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