find where the projectile is level with the cliff.Could someone show me how to approach this question and other similar projectile motion questions
oops, slight error with the t value, but methodology is fine. 10sin30t value is t + time to that point. rhs t value is just time from that point.find where the projectile is level with the cliff.
then, s = ut+1/2 at^2
50 - 10sin30t = ut+1/2at^2 (y-axis)
u get value of t from the point where the projectile is level with the cliff. v(x)t + distance of level with cliff point to cliff..
but that question is kinda retarded. since technically, the object could slide back down the slope if friction is not big enough.
ngl i made my answer way more complicated than it needs to be lol.oops, slight error with the t value, but methodology is fine. 10sin30t value is t + time to that point. rhs t value is just time from that point.
in case anyone doesn't understand where he got the second formula from. That's the equation path formula from 3 unit maths projectile motion and this equation can also be obtained from known physics, albeit after a few steps.tan(30) = 1/sqrt(3)= opp/adj = rise/run
Setting the origin at the cliff the slope is given by (1/sqrt(3))*x -50 = y
The motion of the projectile is given by
(-g*x^2)/100 + x = y
Solve for x:
(1/sqrt(3))*x -50 = (-g*x^2)/100 + x
By applying the quadratic formula you will answer.
Remember to write down why one of the solutions was discarded.