Question

A man throws a ball to maximum horizontal distance of 80 m.calculate the maximum height reached. Plz fast

Answer 1

Answer:

Explanation:

Hope it helps

Answer 2

Given:

• Max horizontal range is 80 metres.

To find:

• Maximum height attained by object?

Calculation:

This is a case of Ground-Ground Projectile.

• We know that for maximum horizontal range, the angle of projection has to be 45°.

$R_{max} = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies 80= \dfrac{ {u}^{2} \sin(2 \times {45}^{ \circ} ) }{g}$

$\implies 80= \dfrac{ {u}^{2} \sin( {90}^{ \circ} ) }{g}$

$\implies 80= \dfrac{ {u}^{2} }{g}$

Now, max height be H :

$H = \dfrac{ {u}^{2} { \sin}^{2} ( \theta) }{2g}$

$\implies \:H = \dfrac{ {u}^{2} { \sin}^{2} ( {45}^{ \circ} ) }{2g}$

$\implies \:H = \dfrac{ {u}^{2} { ( \dfrac{1}{ \sqrt{2} }) }^{2} }{2g}$

$\implies \:H = \dfrac{ {u}^{2} }{4g}$

$\implies \:H = \dfrac{1}{4} \times \dfrac{ {u}^{2} }{g}$

$\implies \:H = \dfrac{1}{4} \times 80$

$\implies \:H = 20 \: metres$

So, maximum height attained is 20 metres.