Need help with a projectile up an incline problem

[daniel1919]

Homework Statement

A projectile is launched at an angle 45degrees with respect to the horizontal and up a hill that is inclined 20degrees to the horizontal. The initial speed of the projectile is 50m/s. Neglect air resistance.
a)what is maximum height? - I found 63.8m which is correct
b) what is the speed of projectile 2.5s after launching? - 37m/s which is correct
c) find the distance up the hill the projectile lands.
d)calculate the magnitude and direction of the velocity of the projectile at impact.

Homework Equations

for part c
x=vicos(45)t-.5gsin(25)t^2
x=visin(45)t-.5gcos(25)t^2

The Attempt at a Solution

The only idea i have for part c is to change the x-y axis so that x is along the slope. Any help solving or ideas would be great.

 

[tiny-tim]

Welcome to PF!

Hi daniel1919! Welcome to PF!

(have a degree: º and try using the X² and X₂ tags just above the Reply box )

c) find the distance up the hill the projectile lands.
…
x=vicos(45)t-.5gsin(25)t^2
x=visin(45)t-.5gcos(25)t^2

(i assume one of those is y ? )

ok, you have the equations for axes along and perpendicular to the slope …

now put y = 0 to get t, and then substitute into x …

what do you get? ​

 

[zgozvrm]

The projectile takes a parabolic path. What is the equation of that path (forget about the hill, for now)?

The hill describes a line; what is it's formula?

Assume the origin to be the point of launch which lies on both the parabola and the hill.
Find the other intersection.

 

[inky]

Homework Statement

A projectile is launched at an angle 45degrees with respect to the horizontal and up a hill that is inclined 20degrees to the horizontal. The initial speed of the projectile is 50m/s. Neglect air resistance.
a)what is maximum height? - I found 63.8m which is correct
b) what is the speed of projectile 2.5s after launching? - 37m/s which is correct
c) find the distance up the hill the projectile lands.
d)calculate the magnitude and direction of the velocity of the projectile at impact.

Homework Equations

for part c
x=vicos(45)t-.5gsin(25)t^2
x=visin(45)t-.5gcos(25)t^2

The Attempt at a Solution

The only idea i have for part c is to change the x-y axis so that x is along the slope. Any help solving or ideas would be great.

For (c)
You need to follow tiny tim method. Hello Tiny tim is this correct.
For vertical axis
y=0
0=(50cos45)t-(0.5 gcos20)t^2
t=7.525 s
range=(50cos65)*7.525/cos20 =169.21 m

 

[zgozvrm]

That's not what I came up with...

First of all we have:

V_i = 50 m/s (initial velocity)
\alpha = 45^\circ (angle of trajectory)
\beta = 20^\circ (angle of the hill)Therefore,

V_{ix} = V_i \times cos(\alpha) = V_i \times cos(45^\circ) (initial horizontal velocity)
V_{iy} = V_i \times sin(\alpha) = V_i \times sin(45^\circ) (initial vertical velocity)but since cos(45^\circ) = sin(45^\circ)[/tex], we have<br /> <br /> V_{ix} = V_{iy}We know that<br /> <br /> t = \frac{X}{V_{ix}}<br /> <br /> and<br /> <br /> Y = V_{iy} \cdot t - 4.9t^2<br /> <br /> so<br /> <br /> Y = V_{iy}\cdot \frac{X}{V_{ix}} - 4.9\left(\frac{X^2}{V_{ix}^2}\right)<br /> <br /> but since V_{iy} = V_{ix}[/tex], we have&lt;br /&gt; &lt;br /&gt; Y = X - 4.9 \cdot \left( \frac{X^2}{V_{ix}^2} \right) (equation A)The slope of the hill is m = tan(\beta) = tan(20^\circ)[/tex] so the equation for the hill is&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Y = X \cdot tan(20^\circ)Substituting for Y in the parabola equation (equation A) gives us&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; X \cdot tan(20^\circ) = X - 4.9 \cdot \left( \frac{X^2}{V_{ix}^2} \right)&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; tan(20^\circ) = 1 - 4.9 \cdot \left( \frac{X}{V_{ix}^2} \right)&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; \left(\frac{4.9}{V_{ix}^2}\right) \cdot X = 1 - tan(20^\circ)&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; X = \frac{V_{ix}^2(1 - tan(20^\circ))}{4.9} = \frac{(50 \cdot cos(45^\circ))^2 \cdot (1 - tan(20^\circ))}{4.9}&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; X = \frac{35.355^2 \cdot 0.636}{4.9} = \frac{1250 \cdot 0.636}{4.9} = \frac{795.037}{4.9} = 162.252Plug this value back into equation A, above, and you get Y = 59.055&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; The distance up the his is found using the Pythagorean Theorem:&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; D^2 = X^2 + Y^2 = 162.252^2 + 59.055^2 = 29813.373&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; and D = 172.665 m&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; (note: I only rounded off in the text, not in my actual calculations)