Given:
- Initial speed of arrow is 15 m/s.
- Angle of projection = 30°.
To find:
- Velocity at highest point?
Calculation:
- This is an example of GROUND-GROUND PROJECTILE.
We know that , acceleration due to gravity is applicable only along Y axis , hence the velocity component along X axis will always remain constant.
At highest point, the projectile only has the X component velocity (because Y component becomes zero).
[tex]v_{highest} = v \cos( \theta) [/tex]
[tex] \implies v_{highest} = 15 \times \cos( {30}^{ \circ} ) [/tex]
[tex] \implies v_{highest} = 15 \times \dfrac{ \sqrt{3} }{2} [/tex]
[tex] \implies v_{highest} = 12.99 \: m {s}^{ - 1} [/tex]
So, velocity at max height is 12.99 m/s