A ball thrown from the edge of a building hits the ground at an angle of 60°with the horizontal, 25 m from the building and 2.5s after it is thrown. (take g = 10m/s 2 and 
= 17)
The direction of velocity with the horizontal with which the ball was thrown, is
Text Solution
Verified by ExpertsA
For horizontal motion, as component of velocity in this direction is not changed with time
t = 2.5 s = 25m/u x ⇒ ⇒ u x =10 m/s
Also at the moment of landing, velocity is V, then
V cos 60 o = u x ⇒ ⇒ V = 
= 20 m/s
and V y = V sin 60 0 = 
= 
m/s
For the vertical motion, using v = u + at and taking upward direction as positive
- 
= u y + (-10)(2.5)
⇒ ⇒ u y = 25 - 
= 8 m/s, upward.
Angle of projection tan θ θ = 
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