Bear with my notation.bazookie said:how did you get that equation?
philxe said:you can use the maths formulae but is it true that they will only award 1 mark for the answer since the working isn't following the physics formulaes.
Anyway, you solve this prove by using the horizontal equation and vertical equation.
using 7 = V cos 15 x T, you derive an equation for T
then substituting T in to 0 = V sin 15 x T - 4.9 T^2
and you should come up with V^2 = 240.1 / 7 tan 15 x cos^2 15
taking the positive square root giving you 11.7 approx
i don't get it, if i use that formula then what is the value for theta? do i add the angle of the projectile to 20? if i do that then d will be 8 (for the Q above). PLEASE HELP!Forbidden. said:http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
Proof it is 11.7 m/s
Use this equation:
maths is over... dont pollute BOS with useless CRAPbazookie said:a long jumper left the ground at an angle of 15 degrees and jumped a distance of 7m. Considering this result, calculate the velocity of the long jumper when leaving the ground.
answer is 11.7
how the fuck?
Such violence, I suggest go for a paddleboat ride.bazookie said:yea this is physics, fuckstick. How is it useless crap when i'll be needing this in two days?
GTFO, stop polluting bos with YOUR useless crap.
edit: fuck you kthnxbye
Actually, theoretically it's maths , but yeah i get your point lol.bazookie said:yea this is physics, fuckstick. How is it useless crap when i'll be needing this in two days?
GTFO, stop polluting bos with YOUR useless crap.
edit: fuck you kthnxbye