Question

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

Answer 1

max range is for angle 45
$80 = \frac{ {u}^{2} \sin^{2}( a ) }{g} \\ u = 40$
max height=
$\frac{ {u}^{2} { \sin }^{2} 2 \times 45}{2g}$
h=80/2m=40m

Answer 2

$maximum \: range = \frac{ {u}^{2} }{g} \\ maximum \: height = \frac{ {u}^{2} }{2g} \\maximum \: range \: is \: given \: so.. \\ 80 = \frac{ {u}^{2} }{g} \\ takng \: g = 10m {sec}^{ - 2} \\ {u}^{2} = 800 \\ \\ now \: put \: {u}^{2} \: in \: maximum \: height \: formula \: so \\ \\ max.height = \frac{ {u}^{2} }{2g} \\ = \frac{800}{20} \\ max.height = 40m$
hope it will help you!