Heidi T. answered • 23d

MA in Mathematics, PhD in Physics with 7+ years teaching experience

The direction of motion of Snowball A is downward or -90 degrees wrt the horizontal. Not sure if this is what you meant by "zero"

To find the direction of motion of Snowball B is found by solving the equations of motion. Since the Define the positive y direction as up (since moving up originally) and positive x direction in the direction of the horizontal motion. I suggest drawing a picture to make it more clear.

Keep in mind, since the acceleration is only in the y direction, the velocity in the x-direction remains unchanged. If you have your picture drawn, you can break the velocity vector into its components - v_{0x} = v_{0} * cos(35^{o}) and v_{0y} = v_{0} * sin(35^{o}) (because the angle was given wrt the horizontal)

The equations of motion in y are:

y = y_{0} + v_{0y}*t + (1/2) a_{y} * t^{2}

v_{y} = v_{0y} + a_{y} * t

where y_{0} = 9 m, y = 0 (at the ground), v_{0y} = v_{0} * sin(35^{o}) = (18 m/s)* sin(35^{o}) = 10.3 m/s, and a_{y} = -9.8 m/s^{2 } (the acceleration due to gravity operating in the negative direction as defined above) . Use the quadratic formula to solve for t.

t = {-(10.3 m/s) +/- √[(10.3 m/s)^{2} - (4)(-4.9 m/s^{2})(9 m)]} / [2*( -4.9 m/s^{2})] use the negative root only since time must be positive (and there is a negative in the denominator)

t = {(10.3 m/s) + √[(10.3 m/s)^{2} + (4)(4.9 m/s^{2})(9 m)]} / [2*( 4.9 m/s^{2})] (factor out a -1 from numerator and denominator then cancel, to make all values positive)

t = 2.77 s

Use the time calculated to find the v_{y} at the end of the journey:

v_{y} = v_{0y} + a_{y} * t = (18 m/s)* sin(35^{o}) - (9.8 m/s^{2})*(2.77 s) = 10.3 m/s - 27.1 m/s = -16.8 m/s

The velocity is negative, indicating the direction is down.

The velocity in the x-direction is unchanged, v_{0x} = v_{0} * cos(35^{o}) = (18 m/s) cos(35^{o}) = 14.7 m/s

These are the components of the velocity vector just prior to Snowball B impacting the ground. If you draw the vector diagram, you will have a right angle below the horizontal. Find the angle of the velocity vector wrt the horizontal using the inverse tangent function. Recall, tanθ = opp /adj, and θ = tan^{-1}(opp/adj)

in this case, opp is v_{y} and adj = v_{x}, --> θ = tan^{-1}(-16.8/14.7) = -48.8 which means the direction is 48.8 degrees below the horizontal.