M2 help plzWatch
i dnt understand how 2 calculate the angle???
v^2 = u^2 + 2as
v = root[(-10 x 3/5)^2 + 2 x 9.8 x 26.1]
v = 23.4
The horizontal component of the velocity is constant, ie 10cos alpha = 8
tan beta = 23.4/8
beta = 71 degrees.
Draw a triangle to help you see that, if it helps. The vertical component on the opposite side and the horizontal on the adjacent side. Hope that makes sense.
i assume you got a) and b) correct?? Anyway at A the particle has horizontal velociry of 3x10 [email protected] = 8m/s, and vertical velocity of 10 [email protected] - 3g = 23.4m/s
Now the tangent of the angle created by the direction of motion is 23.4 / 8 = 2.925. So the angle is arctan 2.925 which is 71 degrees, the answer in the back.
i was just doing it and someone else already posted it damn it.