Physics question regarding kinematics in 2-dim.?

Question

Suppose a stone is thrown at an angle of 30.0 degrees below the horizontal. If it strikes the ground 57.0 m away, find (a) the time of flight. (b) the initial speed, (c) the speed and the angle of the velocity vector with respect to the horizontal at impact.

I know the answers are: (a) 1.57s, (b) 41.9m/s, (c) 51.3m/s, -45.0 degrees.

I just don't know how to start the problem b/c of the fact that the stone was thrown below the horizontal.

Answer 1

You solve it exactly as you would if it was 45 degrees above the horizontal. First you must make an assumption that you are throwing the stone from some initial height y. You must also realize that the path of the objects motion is a straight line and become the hypotenuse of a right triangle. Therefore y is given by: y=57/tan(45)

From Newtons second law, the equations of motion are:
X(t)=vcos(45)t --> 56=vsin(45)t
Y(t)=-.5gt^2-vsin(45)t+y --> 0=-.5gt^2-vsin(45)t+57/tan(45)

So you have 2 equations with 2 unknowns. Solve for v and t simultaneously to get the answers to a and b.

Once you have those answers part c should be very straight forward. Just plug t=1.57 and v= 41.9 into the x and y component equations for velocity solving for those x and y components. Then use the Pythagorean theorem to find the magnitude of the velocity vector.

x: V(t)=vcos(45)
y: V(t)=-.5gt^2-vsin(45)

Good luck...hope this was clear. If not drop me and email.

Answer 2

Hi. Imagine throwing the stone downward from a cliff. Instead of an initial velocity upwards it simply has an initial velocity downwards. Does this help?

Answer 3

assume constant velocity distance by a= 15* t distance by b= 20* t t must be constant as they meet simultaneously 15*t + 20*t =1000 meters find t put value in distance equation find it