Projectiles

A ball B is projected with speed V at an angle θ above the horizontal from a point O on horizontal ground. The greatest height of B above O is H and the horizontal range of B is R. The ball is modelled as a particle moving freely under gravity.
Show that
1. H= (Vsinθ)^2 / 2g
2. R=(V^2)(Sin2θ)/g

To get you started with (1), perhaps try applying the standard SUVAT equation v^2 = u^2 + 2as to the vertical motion of the ball. The vertical speed of the ball will be zero when it is at its highest point. Post your working if stuck.

Original post by Joseph McMahon

See attached for derivation.

Please see the maths forum posting guidelines (sticky). Hints not full solutions. Please amend your post accordingly.

Original post by old_engineer

Please see the maths forum posting guidelines (sticky). Hints not full solutions. Please amend your post accordingly.

"Make sure that the student has attempted the question and posted their working before helping."

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