ok so i plugged 3.9 m and 0m/s into the displacement equation so it makes t=.892s
and then i found out Vi for vertical is 8.74 m/s
then i used SOH to find the hyponuse to be equal to 17.48 by dividing 8.74 by sin30
is this the right answer?
i know to do that.. but i still don't get how u can use the equations when u only have the horizontal displacement of 7.80 m ? how do i use the 30 degree angle to find anything?
this is easy
they give u the angle and initial velocity so u can find out initial velocities for both vertical and horizontal components by multiplying 20*cos30 and 20*sin 30
then u can use an equation to find the time and then use another equation to get the displacement
simple stuff really...
hey guys imm new here
i got a question in my book i can't figure out
it says:
an athlete executing a long jump leaves the ground at a 30 degree angle and travels
7.80m (a) what was the takeoff speed? (b) if this speed were increased just 5.0 % how much longer would the jump be?
anyone...