8/18/2019 Physics Formula notes

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8/18/2019 Physics Formula notes

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Projectiles (1)Object projected vertically upwards1. Maximum height reached (H) = ½ gt = u!g

. "t a#y other height except the maximum$h = ut %½gtspace v = u %gh spa where g = & '. ms&

Object projected horio#tally

*he horio#tal dista#ce travelled (s) = horio#tal velocity x time = vxt = ut

*he vertical dista#ce travelled (h) = uyt % ½ (gt) = ½ (gt) si#ce uy = +

,elocity a-ter a time t$ space v = (vx % vy

)1!

irectio# o- motio# a-ter time t$ space ta#θ = vy!vx/0uatio# -or the path$ spaceh = ½ (gt) = ½ (gs!u) = s( ½ )(g!u)

 

Projectiles (2)Object projected at an angle A to the horizontal(a) RangeRange (R) = horizontal velocity x time = !cos A x 2! sin A"#g =!22 sin A cos A"#g

Range (R) = !2sin 2A#g

$he maxim!m range %or a given velocity o% projection &ill be &hensin 2A = 1' that is' &hen 2A = o or &hen A = *+o

(b) ,eight$he projectile &ill reach its maxim!m height &hen the vertical

8/18/2019 Physics Formula notes

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component o% its velocity is zero' that is' &hen ! sin A - gt = ' or&hen t = ! sin A#g"

.axim!m height (,) =

!2sin2A"#2g