- #1
leah3000
- 43
- 0
referring to an object being projected upwards at an angle ө
i'm a phys student but i don't do math so I'm having trouble understanding how this equation has been simplified.
Horizontal Range= vcosө x 2vsinө/g
=v^2 2sinөcosө/g
this is the line i don't understand : v^2 sin^2 ө/g
where did the cosө go? is is a trig identity that cancels?
also when considering upward motion is g taken as negative?
if so, then why is the time taken to reach the maximum pt given by:
t= v sinө/ g wouldn't it be v sinө/ -g
isn't that the reason for the vertical distance traveled being given by;
y= v sinө t- 1/2 gt^2 ??
i'm a phys student but i don't do math so I'm having trouble understanding how this equation has been simplified.
Horizontal Range= vcosө x 2vsinө/g
=v^2 2sinөcosө/g
this is the line i don't understand : v^2 sin^2 ө/g
where did the cosө go? is is a trig identity that cancels?
also when considering upward motion is g taken as negative?
if so, then why is the time taken to reach the maximum pt given by:
t= v sinө/ g wouldn't it be v sinө/ -g
isn't that the reason for the vertical distance traveled being given by;
y= v sinө t- 1/2 gt^2 ??
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