Question

a cricketer can throw a ball to maximum horizontal distance of 160 M calculate the maximum vertical height to which he can throw the ball

Answer 1

So we know that R or range = 160m

We know apply the formula for calculating range of projectile motion, where "u" is initial velocity, "g" is acceleration due to gravity and "alpha" is the angle at which the ball is thrown.

For maximum horizontal distance, the cricketer must throw the ball at an angle of 45°.

$\frac{ {u}^{2} \sin(2 \alpha ) }{g} = 160$
We now substitute the value of alpha in our equation.

$\frac{ {u}^{2} \sin(90) }{g} = 160 \\ \frac{ {u}^{2}}{g} = 160$
We know rearrange the following equation of motion in the given way with appropriate signs, where "v" if the final velocity, "u" and "g" are the same as before and "s" is the distance travelled.

${v}^{2} - {u}^{2} = 2( - g)s \\ - {u}^{2} = - 2gs \\ \frac{ {u}^{2} }{2g} = s$
And we substitute the value from the previous equation to get the value of "s", which is now the maximum vertical distance which we had to find.

$\frac{160}{2} = s \\ s = 80$
Hence the answer is 80m.

Answer 2

$\bold{Heya......!!!}$

In the question it's given that , the cricketer can throw a ball maximum Horizontal Distance of 160 m .

We have to take out maximum vertical height.

Range. = 160 m

$\bold{Formulae:}$

$\huge\boxed{R = u^2sin2\beta/g}$

If , β = 45° then it's Range is maximum .

→ Maximum Range = $\bold{R = u^2/g}$

→ $\bold{R = u^2/g}$ = 160

=> from the above , we can take out the value of " u "

→ u^2 = 160*10
→ u^2 = 1600 = $\sqrt\bold{1600}$

→ $\bold{u = 40m/sec}$

Now , we know that maximum vertical height ( H ) is :-

$\huge\boxed{H = u^2/2g}$

putting the value of ' u ' , we get

→ $\bold{(40)^2/(2*10)}$

→ $\huge\boxed{H = 80m}$

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Hope It Helps You. ☺