Projectile Motion (1 Viewer)

pomsky · Feb 21, 2015

Hwello c: Could you guys please give working out? You have my eternal gratitude <3

'A ball is thrown with initial velocity 40m/s and angle of inclination 30 degrees from the top of a stand 25 metres above the ground. What is the impact speed and angle?"

Ans: v= 10sqrt(21) m/s and theta = 40^o54'
Why can't we find impact speed through distance/ time? And what relationship does impact speed have to the x and y velocities, i.e why is the relationship pythagoras?

pomsky · Feb 21, 2015

Bumpity-bump ( `w`)/

InteGrand · Feb 21, 2015

pomsky said:
Hwello c: Could you guys please give working out? You have my eternal gratitude <3

'A ball is thrown with initial velocity 40m/s and angle of inclination 30 degrees from the top of a stand 25 metres above the ground. What is the impact speed and angle?"

Ans: v= 10sqrt(21) m/s and theta = 40^o54'
Why can't we find impact speed through distance/ time? And what relationship does impact speed have to the x and y velocities, i.e why is the relationship pythagoras?

Impact speed means how fast its going in the direction its travelling, whereas the x and y components of velocity are how fast its going in the x and y
directions. If you draw a triangle, you'll see that the speed in the direction of travel at any given time is always $v = \sqrt{v_x ^2 + v_y ^2}$ , where v_x and v_y are the x and y components of the velocity at any given point in time.

For the given question, here's a way to do it:

- Set up the equations of motion as usual (x and y as functions of t)
- Find the time t₁ such that the projectile hits the ground (i.e. y = 0, if you take y = 0 as the ground level)
- Sub. in t₁ to your equations for $\dot{x} (t)$ and $\dot{y} (t)$ and then calculate speed using Pythagorean theorem
- To find the angle, note that $\frac{\text{d}y}{\text{d}x}=\frac{\frac{\text{d}y}{\text{d}t}}{\frac{\text{d}x}{\text{d}t}}= \frac{\dot{y}}{\dot{x}}$ . Use this to find the slope of the tangent of the parabola of flight trajectory at the time t₁ (which is when the projectile hits the ground), and then calculate angle from this.

pomsky · Feb 21, 2015

Woah, thanks again Integrand!

Another question (I never seem to run out, huh? haha) how would you find range if you knew max height, but not theta? (theta being the angle in ucos(theta) and usin(theta)) U= 8m/s and g= 10 m/s^-2 . Is g always -10?

"A long-jumper running at 8m/s jumps into the air, rising to a height of 0.6m. What is the length of his jump assuming g= 10ms^-2?"
Ans: 5.00m

pomsky · Feb 22, 2015

bummp~

pomsky · Feb 22, 2015

pew pew.
bump.

InteGrand · Feb 22, 2015

pomsky said:
Woah, thanks again Integrand!

Another question (I never seem to run out, huh? haha) how would you find range if you knew max height, but not theta? (theta being the angle in ucos(theta) and usin(theta)) U= 8m/s and g= 10 m/s^-2 . Is g always -10?

"A long-jumper running at 8m/s jumps into the air, rising to a height of 0.6m. What is the length of his jump assuming g= 10ms^-2?"
Ans: 5.00m

g is the local acceleration due to gravity. On Earth, g is about 9.8 m s^-2, which is approximately 10 m s^-2. If you're asked to calculate something numerically and the question doesn't tell you what to use for g, use either 9.8 or 10 (10 would be easier to work with), and say what value for g you're using.

Here's how to do the question you posted.

Let the angle of the jumper's jump with the ground be $\theta$ . We don't know what this is yet, but it's some constant value between 0 and 90 degrees which we will be able to find using the given info. The initial jump speed was 8 m s^-1, so the x and y components of the initial speed are $8\cos \theta$ and $8\sin \theta$ respectively.

Now, set up your usual equations of motion (I assume you know how to do these), the relevant ones will be:

$x=(8\cos\theta)t \text{ }(1)$

$\dot{y}=-gt+8\sin \theta \text{ }(2)$

$y=-\frac{g}{2}t^2+(8\sin \theta)t \text{ }(3)$ .

Now we're given that the max. height is 0.6 m. Max. height occurs when $\dot{y}$ is 0 (e.g. when you throw a ball in the air, when it reaches its maximum height, it just stops moving in the vertical direction). But $\dot{y}=0$ when $-gt+8\sin\theta = 0 \Rightarrow t=\frac{8\sin \theta}{g}$ .

So at this time, y = 0.6, i.e. $-\frac{g}{2}\left ( \frac{8\sin\theta}{g} \right )^2+(8\sin \theta)\cdot\frac{8\sin\theta}{g}=0.6$ (from Eqn. 3)

$\Rightarrow \frac{32\sin^2 \theta}{g}=0.6$

$\Rightarrow \sin^2 \theta=\frac{0.6g}{32}=\frac{3g}{160}$ .

So $\sin \theta = \sqrt{\frac{3g}{160}}$ . Now we've found $\sin \theta$ (previously an unknown) in terms of known values.

Now, at this time, x is half the range, since when the max. height is reached, you're at half the range, by symmetry of a parabola.

At this time, from Eqn. 1,

$x=8\cos \theta \times \frac{8\sin \theta}{g}$ (since $t=\frac{8\sin \theta}{g}$ ).

But $\cos \theta = \sqrt{1-\sin^2 \theta}=\sqrt{1-\frac{3g}{160}}$ , so at this time,

$x=64\sqrt{1-\frac{3g}{160}} \times \sqrt{\frac{3g}{160}} \times \frac{1}{g}$ .

This is half the range, so the full range is double this, i.e.

$\text{range} = 128\sqrt{1-\frac{3g}{160}} \times \sqrt{\frac{3g}{160}} \times \frac{1}{g}$ .

Plug in g = 10 m s^-2, and you'll get the range to be 4.995998399....m, or 5.00 m to 3 sig. fig.

Projectile Motion (1 Viewer)

pomsky

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InteGrand

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