Physics Angle Velocity?

Question

Ok say a car rolled down a hill that was at an angle of 30 degrees and reached a final velocity of 40mps, the same car repeated this at 40 degrees and reached 49mps, what formula/ math functions, can be used to show the relation between the angle and the speed

Answer 1

well, mgSinX - UmgCosX is the force acting on the car,
where X is the angle. That's the parallel component of gravity on the car - the perpendicular (reaction) times the coeffient of friction( u is a mu) SO:
f=ma
a=f/m
so a =g(SinX - UCosX)
so!
v=u+at assuming u =0( starts at rest)
v=g(SinX - UCosX)t
where t is the time it's rolling for.
If there is no friction u=0.

Answer 2

f = ma = W sin(theta) - k W sin(theta) = W sin(theta)(1 - k); where W = mg, the weight of the car of mass m and g acceleration ~ 10 m/sec^2 on Earth's surface. [Note: If you discount friction, let k = 0, the friction coefficient in the above eqn.] a = the acceleration of the car along the slope of the hill. f is the net force acting on the car (e.g., component of weight along the hill, minus the friction force holding it back).

v^2 = u^2 + 2aS; where v = 40,49 mps the bottom out velocities, u = 0 the starting velocity in each case (presumed), a = f/m = mg sin(theta)(1 - k)/m = g sin(theta)(1 - k) from the first paragraph.

Thus, v^2 = 2g sin(theta)(1 - k)S; where S is the distance the car rolled along the hill. If we let v = 40 mps and V = 49 mps, we can do a ratio (V/v)^2 = sin(THETA)/sin(theta); so that V^2 = v^2 sin(THETA)/sin(theta); where THETA = 40 deg and theta = 30 deg. This shows that, on a relative basis, the velocities vary as the square root of the sine ratio; i.e., V = v sqrt(sin(THETA)/sin(theta))

By the way, your velocities V and v do not check out using the sqrt sine ratio just derived. According to v = 40 mps, V should = 45+ mps; not 49 mps.

Answer 3

If the boy drops the ball to loose fall, ultimate right here would desire to practice preliminary speed is 0 m/s very very final speed is 14.7 m/s Acceleration is (gravity =9.8 m/s2) distance = a million/2 * gt^2 = a million/2 * 9.8 * 4 = 19.6 m If the boy ability throws the ball with preliminary speed of 14.7 m/s, then ultimate right here would desire to practice d = Vi *t + a million/2 gt^2 d= 14.7*2 + a million/2*9.8*4 = 40 9 m