2 balls thrown, what is the speed

jensson

Homework Statement

Two balls are thrown with equal speeds from the top of a cliff of height h. One ball is thrown at an angle of alpha above the horizontal. The other ball is thrown at an angle of beta below the horizontal. Show that each ball strikes the ground with the same speed, and find that speed in terms of h and the initial speed. (Ignore any effects due to air resistance.)

The Attempt at a Solution

I have tried all different ways but I have no idea what is correct. I have equations (well, examples really) for when something is thrown above a horizontal, but not below it.

I have a ridiculously complicated answer and I feel like this should have a simpler answer.

Can anyone offer any help?? a point in the right direction?

The Attempt at a Solution

StephenP
Are you familiar with the work-energy equations? Your life will be much simpler if you try equating the initial (potential and kinetic) energy with the final (solely kinetic) energy.

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Speed =$$\sqrt{{v_x}^2+{v_y}^2}\,.$$

One way to do it is to use conservation of energy.

jensson
unfortunately I am not familiar with work-energy equations... or conservation of energy.

StephenP
In that case, you can use your kinematics equations. For the ball which is thrown downward, you can resolve the initial velocity into its x- and y-components. Then, you have the initial speed, the acceleration, and the distance it falls. From this you should be able to calculate the final speed.
For the ball thrown upward, you again have the initial speed and acceleration, but will have to figure out how far it goes up before it starts free-falling.
Good luck!

jensson
for the ball thrown downward,
I know the x and y components are

initial v cos theta
and
initial v sin theta

so if I'm solving for the speed, would I have

V(for x) = initial v cos theta
and
V(for y) = initial v sin theta - gt

for distance:
x = h + (init v cos theta) (t)
y = h + (init v sign theta) (t) - 1/2 (g)(t^2)

acceleration in the x direction is 0 and in y it is gravity?

I think trying to work this problem out I have confused myself behind repair. Am I making any sense with my answers?

StephenP
In your problem statement, you said you are only trying to find the speed with which it lands. So, for the ball being thrown down, don't worry too much about calculating the distance.
What we know is that the cliff is at a height h, so the y-distance is going to be only h.
You have the initial y-velocity correct. Now, we need to figure out by how much the x- and y-velocities will change during the fall of distance h.
Remember that there is no acceleration in the x-direction.
For the y-direction, use the basic kinematics equation $$\vec{v_f}^2=\vec{v_i}^2+2\vec{a}d$$. Once you have the final x- and y- velocities, use the equation that SammyS posted above to re-combine them into the total final velocity.