Question

A body is projected up with 3/4th of the escape velocity from the earth's surface.the height reached by the body is

Answer 1

$\frac{3}{4} \times 11.2 \: \frac{km}{s}$
u =8.4 km/s
v=0
s=?
a= -9.8 m/s^2 =0.0098 km /s^2
$now \: {v}^{2} = {u}^{2} \times 2as$
by solving ,1.382976 km (approx)