Two identical objects start at the same height above level ground. Simultaneously, object A is launched at an angle of 30° above horizontal and object B is launched at angle of 30° below horizontal.

Question

Two identical objects start at the same height above level ground. Simultaneously, object A is launched at an angle of 30° above the horizontal and object B is launched at an angle of 30° below the horizontal. Both objects have a launch speed of 20 m/s. Assume that air resistance is negligible.

a) Which object hits the ground first, or do they hit the ground at the same time?

b) Which object hits the ground with the greatest speed, or do they hit the ground with the same speed?

c) If the two objects started at a height of 5.0 m above the ground, how far apart would they land?

Answer 1

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 * t²

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/s²

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 * t², 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)

Answer 2

Hi Jim

This is an interesting problem.

a) The answer to this question is pretty intuitive. If I throw a ball downward, and another upward, the ball thrown downward will surely hit the ground first, because the upward ball needs to go up and back down before it passes the height from which it is thrown. Meanwhile, the downward ball is only headed down.

b) In the absence of wind resistance, a ball thrown upward, will return to it's starting height at the the same vertical velocity only downward). So when the upward ball passes the starting height it is a mirror image of the ball thrown downward - same vertical and horizontal velocity. From that point on, both balls are on the ame path and controlled by gravity. So, the hit the ground at the identical velocity.

c) The "range" equation tells us how far a ball thrown upward will travel before it returns to the height at which it is thrown. That equation is :

X = v(0) ^2 * Sin (2theta)/g In this case the range is 400sin60/9.8 = 35.35 m

So, the upward guy starts his downward journey 35.35m farther "downfield" than the downward guy. From those two points, they travel the exact same path to the ground. So, they land exactly 35.35 meters apart. It doesn't matter how, high above the ground they started.