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A ball thrown by a boy in the street is caught by another boy on a balcony 4m above the ground and 18m away after 2s.calculate the intial velocity and the angle of projection.

A ball thrown by a boy in the street is caught by another boy on a balcony 4m above the ground  and 18m away after 2s.calculate the intial velocity and the angle of  projection.

Grade:11

1 Answers

Govind Menon
12 Points
7 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 ....
 

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