Projectile motion, finding initial velocity

AI Thread Summary
A rock is thrown from a height of 1 meter at a 35-degree angle, reaching a horizontal distance of 64 meters and a height of 29 meters. The initial velocity is calculated using the equations of motion, with attempts yielding results of 102 m/s, which contradicts the book's answer of 42 m/s. The discrepancy arises from confusion over constants used in the calculations, particularly the incorrect use of 10035.2 instead of 20070.4. Participants emphasize the importance of maintaining symbolic expressions until the final calculation to avoid errors. Clarification and detailed calculations are requested to resolve the differences in results.
fishingspree2
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Homework Statement


A rock is thrown from a 1 meter height with a 35 degrees angle with horizontal. At x=64 m, its height is y=29 m. Find it's initial velocity.


Homework Equations



x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}
y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}

The Attempt at a Solution


Let F = initial velocity
then Vx0 = F cos 35
Vy0 = F sin 35
x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}
64=F\cos \left( 35 \right) t
t=64\,{\frac {1}{F\cos \left( 35 \right) }}

y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}
29=1+64\,\tan \left( 35 \right) - 10035.20000\,{\frac {1}{{F}^{2}<br /> \left( \cos \left( 35 \right) \right) ^{2}}}
solving for F gives 102 m/s
the answer in the book is 42 m/s


sorry i am new to latex
 
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Hi fishingspree2,

fishingspree2 said:

Homework Statement


A rock is thrown from a 1 meter height with a 35 degrees angle with horizontal. At x=64 m, its height is y=29 m. Find it's initial velocity.


Homework Equations



x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}
y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}

The Attempt at a Solution


Let F = initial velocity
then Vx0 = F cos 35
Vy0 = F sin 35
x=x_{{0}}+v_{{{\it x0}}}t+ 0.5\,a_{{x}}{t}^{2}
64=F\cos \left( 35 \right) t
t=64\,{\frac {1}{F\cos \left( 35 \right) }}

y=y_{{0}}+v_{{{\it y0}}}t+ 0.5\,a_{{y}}{t}^{2}
29=1+64\,\tan \left( 35 \right) - 10035.20000\,{\frac {1}{{F}^{2}<br /> \left( \cos \left( 35 \right) \right) ^{2}}}
solving for F gives 102 m/s
the answer in the book is 42 m/s


sorry i am new to latex

The number 10035.2 in the above equation does not look right to me.
 
sorry the calculation was made using 20070.4 i don't know how the 10035.2 showed up

i get 102m/s using 20070.4 and it is not correct according to the book
 
fishingspree2 said:
sorry the calculation was made using 20070.4 i don't know how the 10035.2 showed up

i get 102m/s using 20070.4 and it is not correct according to the book

I'm not getting 102m/s using 20070.4; can you show the details about how you get 102m/s?
 
I sorted out the constant. Thanks alphysicist.

It is usually a good idea to keep your expressions symbolic until the very end.

Show us how you are getting 102m/s.
 
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