Question 1199411
.
A projectile is fired at an angle of 30° to the horizontal with a velocity of 40 m/s. 
Calculate the velocity attained after 1 sec.(g = 10 m/s2)
~~~~~~~~~~~~~~~~~


<pre>
At the starting moment, vertical velocity component is  {{{V[vert](0)]}}} = 40*sin(30°) = {{{40*(1/2)}}} = 20 m/s.


The time to get the maximum height is  {{{t[h_max]}}} = {{{V[vert](0)/g}}} = {{{20/10}}} = 2 seconds,
under the given condition, so after 1 second the projectile is still moving up.



Vertical component of the velocity at t = 1 second is  

    {{{V[vert](1)}}} = {{{V[vert](0) - g*t}}} = 20 - 10*1 = 10 m/s.



Horizontal component of the projectile is  {{{V[hor]}}} = 40*cos(30°) = {{{40*(sqrt(3)/2)}}} m/s
remains constant during the entire flight (till hitting the ground). 



Thus the velocity attained after 1 second is


      V(t=1) = {{{sqrt((V[vert](1))^2 + (V[hor])^2)}}} = {{{sqrt(10^2 + (40*(sqrt(3)/2))^2)}}} = {{{sqrt(100 + 1200)}}} = {{{sqrt(1300)}}} = 36.055 m/s.   <U>ANSWER</U>
</pre>

Solved.