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?
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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1 Answer
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4.16 We know for a projectile motion, horizontal range is given by
R = , where R is the horizontal range, u is the velocity and is the angle of projectile
So = 100/sin 2
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection = 45 , = 100 …….(1)
The ball will achieve max height when it is thrown vertically upward. For such motion, final velocity v = 0
From the equation - =2gH, where acceleration a = -g, we get
0 - = -2gH, H = = = 50m
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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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