Question 1188879

Given,

Angle: {{{45}}}°  

Landing distance: {{{8.2m}}}

We know, the formula to find the range of a projectile is:

{{{R=(u^2sin(2theta))/g}}}

where, {{{u}}} is the initial velocity, {{{theta}}} is the angle of launch and {{{g }}}is the gravitational acceleration 

substituting the values in the formula, we get:

{{{8.2=(u^2sin(2*45))/9.8}}}
{{{8.2*9.8=u^2sin(90)}}}.........sin(90)=1
{{{u^2=80.36}}}
{{{u=8.964(m/s)}}}
Therefore, the takeoff speed is {{{8.964 (m/s)}}}.