Heidi T. answered • 09/30/19

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

To start this problem, draw a picture. I always recommend this; the picture can be labeled and helps to clarify the arrangement. Then define the coordinate system. You may define the coordinate system in any way you like as long as the axes are perpendicular to each other, but it's best to define it in a way that will simplify the problem. I will solve the problem with the positive x in the horizontal direction the objects are launched and positive y is up. The ground is set to zero for the y-direction and zero in the x-direction is the starting point.

(a) Object B hits the ground first. This is because Object A has to go up before it goes down. This is very clear if you draw the picture. Mathematical proof:

Start with the general motion equation: y = y0 + vo * t + (1/2) * a * t^{2}

v0A = 20 sin (30) m/s for object A and voB = 20 sin(-30) = - 20 sin(30) m/s (for the purposes of the proof, all that we need to say is that v0B = - v0A), y = 0, and a = - 9.8 m/s^{2}

Use the quadratic formula to solve for t:

t = { - v0 +/- √[ (v0)^{2} - 4 (a/2) (y0 - y) ] } ⁄ (2 * a ⁄ 2) substitute

tA = {-v0A - √[ (v0A)^{2} + 4 (4.9) (y0) ] } ⁄ (- 9.8) = { v0A + √[ (v0A)^{2} + 4 (4.9) (y0) ] } ⁄ (9.8)

tB = {-(-v0A) - √[ (-v0A)^{2} + 4 (4.9) (y0) ] } ⁄ (- 9.8) = { - v0A + √[ (-v0A)^{2} + 4 (4.9) (y0) ] } ⁄ (9.8)

notice that tA has v0A/9.8 positive and tB has v0A/9.8 negative; this proves that tA is greater than tB You can put in numbers to get the values.

(b) both have the same speed when they hit the ground. This is because the path of A is symmetric, so the speed and direction (vertical) that it started up with is the same magnitude and opposite sign in the vertical when it returns to its starting point. However, it was already established that the initial velocity of object B was same magnitude, opposite sign in the vertical as object A, therefore we can treat Object A as if were launched at the same direction and magnitude as object B, only at a later time. Therefore they will land with the same speed.

Mathematical Proof:

start with general motion equation for velocity: v = v0 + a * t

Recall v0B = - v0A and use the fact that tA = (v0A + [sqrt term]) ⁄ 9.8 and tB = (- v0A + [sqrt term]) ⁄ 9.8

Speed (in y direction) when they hit:

vA = v0A - {(9.8)(v0A + [sqrt term]) ⁄ 9.8 } = v0A - v0A - [sqrt term] = - [sqrt term] (negative indicates downward)

vB = - v0A - {(9.8) (- v0A + [sqrt term]) ⁄ 9.8} = - v0A + v0A - [sqrt term] = - [sqrt term]

Notice I did not need to put in any numbers (could have kept g instead of 9,8) and still was able to prove the objects hit at the same speed. This is because it is a general result. Anytime you have two objects fired with the same speed in opposite directions with respect to the x-axis, their final speeds when they hit the ground will be the same.

(c) How far apart the objects land is determined by the time of flight and the horizontal speeds. Using the equations from part (a) and the value for y0 = 5 m (given) the times of flight for the two objects can be found and used in the equation of motion in the x-direction:

x = x0 + v0 *t + (1/2) a * t^{2}, where x0 = 0, a = 0, v0A = 20 cos(30), v0B = 20 cos(-30) = 20 cos(30) (the horizontal velocities of the objects are the same), and tA and tB are found from the y-direction equations in part (a)

so the equation of motion reduces to

x = v0 * t

Remember, when solving for tA and tB, the velocity is the velocity in the y-direction

tA = {20 sin(30) + √[ (20 sin(30))^{2} + 4 (4.9) (5) ] } ⁄ (9.8) = 2.456 s

tB = {- 20 sin(30) + √[ (20 sin(30))^{2} + 4 (4.9) (5) ] } ⁄ (9.8) = 0.415 s

xA = (20 cos(30) m/s) * (2.456 s) = 42.5 m (43 m)

xB = (20 cos(30) m/s) * (0.415 s) = 7.18 m (7.2 m)

The distance between them is (43 m) - (7.2 m) = 35.8 m (36 m or 35 m, depending on when round)