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Old Mar 31st 2017, 07:05 AM   #1
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Angry Projectile Motion Problem for Beginner Physics Student

I have a problem I can't solve. A stone is thrown from a building 45.0m high at an angle of 30 degree below the horizontal. It strikes the ground 57.0m away.

A. What is the time of flight?
B. What is the initial speed?
C. What is the speed and angle of the velocity vector with respect to the horizon at impact?

The answers are A. 1.57s B. 41.8m/s C. 51.3 and -45.0 degrees.

I CANNOT figure out the time of flight as 1.51s. I've tried all the formulas I know. PLEASE HELP! This is driving me crazy!

Last edited by nackel; Mar 31st 2017 at 09:53 AM.
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Old Mar 31st 2017, 08:43 AM   #2
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Start by setting up equations for x(t) and y(t). First for the horizontal position x(t):

$\displaystyle x(t) = x_0 + v_{0x}t$

The initial horizontal velocity is equal to the stone's initial velocity times cos(30). Let's let T = the time the stone hits the ground - this yields:

$\displaystyle x(T) = 57 = V_0 \cos(30) T$ (equation 1)

Then for the vertical displacement:

$\displaystyle y(t) = y_0 + v_{0y}t + \frac 1 2 a t ^2$

and at t=T:

$\displaystyle y(T) = 0 = 45 - V_0 \sin(30)T -\frac 1 2 g T^2$ (equation 2)

You now have two equations in two unknowns of V_0 and T. Solve for T.
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Old Mar 31st 2017, 09:24 AM   #3
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Forgive me, but I'm still confused. How can I solve for T if I don't know V_o? How can I solve for T if I don't know V_o?
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Old Mar 31st 2017, 09:32 AM   #4
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I used y=-1/2gt(squared) since Vyo=0 and got 3.02s. I just can't figure out what to do to get 1.57s.

Last edited by nackel; Mar 31st 2017 at 09:53 AM.
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Old Mar 31st 2017, 11:21 AM   #5
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Show me your work. If you sub V_0 from the first equation into the second and solve for T it should come out OK. Also, Vyo does NOT equal 0 - it equals -Vo sin(theta). Remember the projectile is thrown at a downward angle, so Vo does have an initial vertical component.
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Old Mar 31st 2017, 02:47 PM   #6
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My work is invalid because I was using 0 for Vyo. I see how right you are about the Vyo existing as the stone is thrown at a 30 degree down angle.

I hope you can put up with my inexperience, but I can't see how to substitute Vo from the 1st equation into the second. I don't have a Vox and I don't know how to find it with only theta and delta x.

This is why the question is driving me crazy. I feel like I don't have enough information to solve the problem, but since I'm new to this, I don't know what to do.
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Old Mar 31st 2017, 06:20 PM   #7
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$\Delta y = v_0 \sin{\theta} \cdot t - \dfrac{1}{2}gt^2$

Using $g = 9.8 \, m/s^2$ ...

$-45=v_0 \sin(-30) \cdot t - 4.9 t^2$

$0=45 + v_0 \cdot \left(-\dfrac{1}{2}\right) \cdot t - 4.9 t^2$

$0=90-v_0 \cdot t - 9.8t^2$


$\Delta x = v_0\cos{\theta} \cdot t$

$57=v_0 \cos(-30) \cdot t$

$57=v_0 \cdot \dfrac{\sqrt{3}}{2} \cdot t \implies v_0 \cdot t = \dfrac{114}{\sqrt{3}}$

substitute $\dfrac{114}{\sqrt{3}}$ for $v_0 \cdot t$ in the $\Delta y$ equation and solve for $t$.

Last edited by skeeter; Apr 1st 2017 at 09:42 AM.
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