Question

A cricketer can throws a ball to maximum horizontal distance of 100m.With same speed how high can he throw the ball.​

Answer 1

He can throw upto 50 m

Given

A cricketer can throws a ball to maximum horizontal distance of 100 m

To Find

How high can he throw the ball ?

Knowledge Required

$\bf R=\dfrac{u^2sin2\theta}{g}\\\\\bf H= \dfrac{u^2sin^2\theta}{g}\\\\\bf t=\dfrac{2usin\theta}{g}$

Maximum range is possible when angle of projection ( θ ) is 45°

Solution

Given ,

• Maximum range , R = 100 m
• so , θ = 45°

Apply formula for range ,

$\bf \red{R=\dfrac{u^2sin2\theta}{g}}\\\\\rm \to 100=\dfrac{u^2sin2(45)}{g}\\\\\to\rm 100=\dfrac{u^2sin90}{g}\\\\\to \bf \dfrac{u^2}{g}=100...(1)$

Apply formula for height ,

$\bf H=\dfrac{u^2sin^2\theta}{g}\\\\\to \rm H=\dfrac{u^2(sin45)^2}{g}\\\\\to \rm H=\dfrac{1}{2}\times \dfrac{u^2}{g}\\\\\to \rm H=\dfrac{1}{2}\times 100$

∵ From (1)

$\to \bf H=50\ m$

So , He can throw upto 50 m

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Answer 2

$\bf{\underline{\underline{Question:-}}}$

A cricketer can throws a ball to maximum horizontal distance of 100m.With same speed how high can he throw the ball.

$\bf{\underline{\underline{Solution:-}}}$

We know ,

• that horizontal range is maximum if the angle of projection is 45° and is given by.

→ R max = $\sf\dfrac{u^2}{g}$

Here, R max = 100 m

so,

→ u² / g = 100 m

If the cricketer throws the ball vertically upward then the ball will attain the maximum height from the ground

H max = u² / 2g

→ 100 /2

→ 50 m.

$\bf{\underline{\underline{Hence:-}}}$

• with the same speed ball can he throw at 50m