- #1

leah3000

- 43

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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 travelled 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 travelled being given by;

y= v sinө t- 1/2 gt^2 ??

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