Question

an object is thrown upwards with a velocity os 10m/s . Find the maximum height reached (g=9.81)

Answer 1

velocity of object =10 m/s

g.( acceleration due to gravity) = 9.81

angle is 90°

so

maximum height =

H=
${v}^{2} \times { \sin( \alpha ) }^{2} \div 2g$
so

H=
$\frac{100 \times \ { \sin(90) }^{2} }{2 \times 9.8}$
=>

$\frac{100}{9.8 \times 2}$


$\frac{100}{19.1}$
=
5 .10 m