Govind Menon
Last Activity: 10 Years ago
Let us first resolute the velocity as ux for velocity along x axis due to which it travels 18m and uy which is vertical velocity due to which it travels 4m.
There is no air drag(assumption)
so, ux=s/t
ux= 9m/s
There is gravitational accn along y axis,
so a=-10m/s^2
by solving this eq ( s= uyt + ½ at^2)
we get uy=12m/s
we know that,
resultant = rt(a^2+ b^2 +2abcos(theta))
so, substituting values,
u= 15m/s
(note: here we have resoluted the initial velocity as ux and uy and the resolutions are perperdicular, so cos(pi/2) = 0
so, u=rt(ux^2+uy^2))
Now , to find the angle,
we use the formula
tan(alpha)=uysin(theta)/(ux+uycos(theta))
where alpha is the angle between the horizontal axis( or x axis, here it is ux) and theta is the angle btw the adding vectors...
if you want angle of projection with respect to y axis then just substitute ux with uy and uy with ux in above eq.)
By solving we get
tan(alpha)= uy/ux
=12/9
=4/3 (because sin90 is 1 ans cos90 is 0)
so alpha = tan^-1(4/3)
this may be kept as final answer but if we further solve,
Angle of projection=42.96 degree.
Hope it helps ....
