Projectile Motion

Question

A ball is launched with an initial velocity of 5 m/s at an unknown angle. After 2 seconds, it returns to the original height when it lands 6 m from where it launched.

A Find the component horizontal velocity.

B Find the component vertical velocity.

A I calculated this with 6 = (Vi + Vf)/2 * 2. Vinitial (x) and Vfinal(x) are the same, so it's 6 = 2V/2 * 2

6 = 2V

3 = Yinitial in the x (horizontal.)

Vinitial y = 4, since it's a 3, 4, 5 right triangle.

This can also be found by using the x component to find the angle: 5 cosx = 3

cosx = 3/5 = .6

x = arccos(.6) = 53.13 degrees.

This is where the trouble comes.

Vinitial y can be found with 5 sin(53.13 degrees) = 4

It can also be found with Vfinal = Vinitial + at

It reaches max height at 1 second, where the Vfinal y = 0.

0 = Vinitial(y) + -9.8*1

9.8 = Vinitial(y)

There are two different answers for Vinitial(y). This isn't correct.

Please help.

Answer 1

The reason there are multiple, contradictory answers is because this problem is overconstrained. There is 1 unknown (the angle of initial velocity), yet there are 2 equations:

The x position after 2 seconds is 6, 2v_x = 6

The y position after 2 seconds is 0, -g*2²/2 + v_y*2 = 0

Since the problem is overconstrained, that means that, unless the solution to the first equation just so happens to equal the solution to the second equation, this problem is unsolvable.

Alternatively, we could phrase it as there being 2 unknowns (the x velocity & the y velocity), and 3 equations:

The x position after 2 seconds is 6

The y position after 2 seconds is 0

The magnitude of the initial velocity is 5m/s, v_x²+v_y² = 5²

Either way, the point still stands that this equation is overconstrained by one degree of freedom.

I have a hypothesis, however, and that is that the person who made this problem meant for the value of g to be an unknown (say, if this takes place on another planet, or if you're on a plane traveling in a parabolic arc).

If that were the case, then we'd have 2 equations and 2 unknowns. Or, 3 equations and 3 unknowns, which, actually, doing it that way makes it easier:

v_x*2 = 6

-g*2²/2 + v_y*2 = 0

v_x² + v_y² = 5²

2v_x = 6, so v_x=3

-g*2²/2 + v_y*2 = 0, -2g+2v_y = 0, v_y = g

v_x²+v_y² = 25, 9+v_y²=25, v_y² = 16, v_y = ±4.

g = v_y = ±4.

Presumably, the gravitational constant is positive.

So, the gravitational constant would be 4 m/s² downwards, and the initial velocity would be 3m/s horizontally and 4m/s upward.

Though technically, it is also possible for the gravitational constant to be negative (for instance, in an aircraft accelerating towards the earth faster than 9.81 m/s², which would simulate negative gravity), in which case the gravitational constant would be 4 m/s² upwards and the vertical component of the velocity would be 4 m/s downwards. However, I'd probably disregard that as a possibility, as it is typically outside the realm of standard physics curricula.

Again, this was just a hypothesis. But as we can see, for this problem to be correct, either the gravitational constant cannot equal 9.81m/s² like it is on Earth, or there must be some other factor that wasn't mentioned in the initial problem.

Since the equation having a different gravitational constant wasn't specified in the initial problem, however, it might not be wise to jump to that conclusion. On the other hand, though, since this problem is unsolvable under the assumption that g=9.81 (or 9.8, or 10, depending on which one they want you to use)...I guess it couldn't hurt? Worst case, you get it wrong. Which you were already going to do, unless you either magically read the teacher's mind and figured out what answer they wanted you to put down, or unless the teacher realized the problem has no solutions and decided not to count that problem.

Answer 2

The problem as posed does not present a valid answer. If you plug in the maximum value for the vertical velocity (5m/s) the time needed to reach the maximum height is about 0.5 sec which is higher then the value it actually takes of 1 sec.

More in detail, the projectile motion can be decomposed as a constant velocity motion along the x axis as no forces act on the ball. On the y axis we have a constant acceleration motion since the force of gravity acts on the ball.

Thus we have:

v_y(t) = v_y⁰ - g t = v⁰ sin(a) - g t

d_y(t) = v_y⁰ t - 1/2 g t^2 = v⁰ t sin(a) - 1/2 g t^2

v_x(t) = v_x⁰ = v⁰ cos(a)

d_x(t) = v_x⁰ t = v⁰ t sin(a)

at any time t. Here a is the angle we are looking for, v_y(t), d_y(t), v_x(t), d_x(t) are the distance travelled the velocity and distance traveled by the ball respectively along the y and x axis. Let's call t₂ the time that it takes for the projectile to reach its maximum height. This is given by imposing v_y(t₂) = 0, which gives

t₂ = v⁰ sin(a)/g

Let's call t₁ the time that it takes for the ball to reach the ground from the maximum height. This is given by imposing that d_y(t₁) as the ball touches the ground at t₁

d_y(t₁) = d_y(t₂) - 1/2 g t₁^2 = v_y⁰ t₂ - 1/2 g t₂^2 - 1/2 g t₁^2 = 1/2 g (t₂^2-t₁^2) = 0 -> t₁ = t₂

which means that the time the ball takes to reach the maximum height is the same it takes to go down.

Thus we have

t = t_{1 +} t₂ -> t₁ = t/2

Then we have a = arcsin(t₂ g/v⁰ ) = arcsin(t g/(2v⁰) ) = arcsin(1 s 9.8 m/s^2 /(5 m/s)) = arcsin(1.96) which does not exist since the sin function is defined between -1 and 1 so the problem does not have a solution. However if the values admitted a solution for the angle that we call a then we would find a with the previous equation and retrieve the values for the horizontal and vertical velocity as

v⁰_x = v⁰ cos(a)

v⁰_y = v⁰ sin(a)