# If an object launches off of a table at a given velocity, calculate angle below the horizontal when it hits the ground? [closed]

So this is a basic kinematics question that I'm having trouble understanding. Here is the question verbatim:

A hockey puck slides off the edge of a table with an initial velocity of 28.0 m/s and experiences no air resistance. The height of the tabletop above the ground is 2.00 m. What is the angle below the horizontal of the velocity of the puck just before it hits the ground?

Here is what I have so far:

Given:

$$v_{0x} = 28 \frac{m}{s}$$ $$v_{0y} = 0 \frac{m}{s}$$ $$y_{0} = 2.00 m$$ $$x_{0} = 0 m$$

I calculated the time the object would be in flight first. From this, I would use the time in flight to calculate the final x-position and use trigonometry to find the angle. However, this did not seem to yield the correct answer:

$$y_{f} = y_{0} + v_{0y}t + \frac{1}{2}at^2$$ $$y_{f} = 0$$ $$0 = 2.00 m + 0 \frac{m}{s} + \frac{1}{2}(-9.8 \frac{m}{s^2})t^2$$ $$4.9 \frac{m}{s^2}t^2 = 2.00 m$$ $$t = \sqrt{\frac{2.00}{4.9}}$$ $$t = 0.64 s$$

Then using time to calculate the final x-position...

$$x = x_{0} + v_{0x}t + \frac{1}{2}at^2$$ $$x = 0 m + (28\frac{m}{s})(0.64 s) + 0$$ $$x = 17.9 m$$

I then constructed a right triangle using the x and y position components and calculated $$\theta$$ using basic trig:

$$tan(\theta) = \frac{2.00}{17.9}$$ $$\theta = 6.38^{\circ}$$

However, the answer to the problem was $$12.6^{\circ}$$. What did I do wrong? Was I correct in using the time it would take the puck to drop from the height tabletop to the ground and then plugging that into the displacement equation for x to create a triangle?

## closed as off-topic by user191954, John Rennie, ZeroTheHero, M. Enns, Kyle KanosDec 3 '18 at 11:07

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• You're doing it wrong- TO CONSIDER A PARABOLA TO BE A TRIANGLE is so much wrong. The path will be parabola, so the tangent at the point will give the angle i.e. Velocity in y direction by velocity in x direction = slope of tangent at any point in parabola. Here, parabola basically is of function x(t). Differentiate to get velocity i.e. tangent. – Quark Feb 12 '16 at 6:26

## 3 Answers This is your problem (not to scale).

The trajectory of the puck is in blue.

The yellow angle is what you're asked to calculate.

The red angle is what you have calculated.

You need to compare the vertical and horizontal velocities, not the vertical and horizontal displacements.

• +1 That is a great answer to such a homework problem! It pinpoints clearly what the problem with OP's attempt is without explicitly giving the whole solution/calculation. – ahemmetter Dec 3 '18 at 8:00

You need to read the problem again. This is a typical problem that beginning physics students have: they don't read the problem enough. You should read a problem at least 3 times before you do any work. Read slowly.

You will see that you're supposed to find the angle of the velocity vector, not the angle of a line between the starting and ending points.

• Not saying you don't have a point... But the velocity vector is horizontal. So how are my calculations incorrect? I read the problem and this is how I understand it. – Kenneth Worden Feb 12 '16 at 6:04
• $v_y = v_{y0} + (-9.8) \sqrt{\frac{2.0}{4.9}}$ – Farcher Feb 12 '16 at 6:58
• If the velocity vector remains horizontal the puck will never hit the floor. There is a downward acceleration because of gravity. Acceleration changes velocity, so the velocity must gain a downward component. Use your time to find the velocity components just before the puck hits the ground. – Bill N Feb 12 '16 at 13:17

We need to understand that, the direction at which an object points to , is the direction of it's velocity at that point, at that time.

Now, the motion of the puck can be viewed as follows:

Imagine hitting the puck horizontally, with no gravity to pull it down. The puck would move smoothly along the horizontal.

Now imagine that we let the puck fall freely in gravity, without giving it any horizontal force. The puck falls with acceleration due to gravity.

Now, we realize that real motion of the puck is the combination of the above two situations!

## Finding the velocity components

The velocity can be split into two components, the horizontal and the vertical.

WE KNOW that there's no force acting horizontally, WHICH MEANS we dont have horizontal acceleration. So we don't have any change in the horizontal component of velocity. $$v_x=28m/s$$

Now, the initial vertical velocity of the object is zero.

Using the familiar equations $v=\sqrt{2.g.h}$, we find the final vertical velocity. $$g=9.8m/s$$ $$h=2m$$ $$v=\sqrt{2.(9.8).2}$$ $$v_y=6.26m/s$$

The angle of the combined velocity with the horizontal is given by: $$\theta=tan^{-1}\frac{v_y}{v_x}$$ Putting the data, $$\theta=12.6^0$$

So, we must take special care of this splitting into component concept. It can be applied in many similar situations when you get stuck!

Hope this clears your doubt!!

• This is a complete solution to a homework-like question. Please do not post complete solutions, and please delete this post. – garyp Oct 8 '17 at 17:41