4.16 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

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    Answered by

    Vishal Baghel | Contributor-Level 10

    4 months ago

    4.16 We know for a projectile motion, horizontal range is given by

    R = u2gsin2θ , where R is the horizontal range, u is the velocity and θ is the angle of projectile

    So u2g = 100/sin 2 θ

    The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection θ = 45 ° , u2g = 100 …….(1)

    The ball will achieve max height when it is thrown vertically upward. For such motion, final velocity v = 0

    From the equation v2 - u2 =2gH, where acceleration a = -g, we get

    0 - u2 = -2gH, H = u22g = 1002 = 50m

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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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