Question

A castle sits at the top of a 100 m tall hill. The slope of the hill is 30 ° with respect to the horizontal. One projectile of mass 10 kg is fired at an angle of 30 ° above the horizontal with a speed of 31. 32 m / s from the top of the hill toward the bottom of the hill. Another projectile of mass 10 kg is fired at the same time at an angle of 60 ° above the horizontal with a speed of 54. 25 m / s from the bottom of the hill toward the top of the hill. Find the point at which they meet.

Answer 1

kinematics equations we will use:

\[r = r_{0} + V_{0}t + \frac{ 1 }{ 2 }at^2\]
and
\[V = V_0 + \frac{ 1 }{ 2 }at^2\]

where r is the position (applicable to both x and y axis)

since the displacement in the x and y directions are independent, we can evaluate them separately, that is, deal with only x first to find out what happens on the axis, the on the y axis, then put them together.

it can also be done simultaneously but this is by far the simplest way. and the way you're expected to do it.

Answer 2

note: even though their masses are the same, in kinematics, mass is meaningless.

particle A:

x axis: X (meeting point) = 0 + 31. 32cos(30) t + 0 [no acceleration in x direction]
y axis: Y (meeting point) = 100 + 31.32sin(30) t + 1/2(-9.81) t^2


particle B:

x axis: X (meeting point) = 173.2 + 54.25cos(120) t + 0 [no acceleration in x direction]
y axis: Y (meeting point) = 0 + 54.25sin(120) t + 1/2(-9.81) t^2

you have 3 unknowns and 4 equations, therefore you can find what you need using your favorite method :P

(turns out we didnt need the other equation, only one of them)

let me know if you need help solving for for t, x, and y.

(find t first)

Answer 3

First projectile (on top):
Vx = vi*cos(30) = (31.32m/s)(cos(30))= 27.12
Vy = vi*sin(30) = (31.32m/s)(sin(30)) = 15.66
--> (27.12, 15.66)

Second projectile (on bottom):
Vx = vi*cos(60) = (54.25m/s)(cos(30))= 27.125
Vy = vi*sin(60) = (54.25m/s)(sin(30)) = 46.98
--> (27.125, 46.98)

I think :]

Answer 4

Oops, nevermind :P

Answer 5

the equation you used, which I said we'd use but didn't, is only good to find the velocities when they meet. it will not find the meting point.

it does work since when you have their velocities when they meet you can find the position for one of them, which is therefore the position for both.

both methods work, except you'd use two equations

Answer 6

one way is to get started is to equate the x axis for particle A to the x axis for particle B and solve for t, then for x.

that same t can be plugged in the y direction for either one. (same y point)

Answer 7

Wow, thank you so much! I wouldn't of thought of using 120°.

A:
X = 0 + 31.32cos(30)t + 0
x = 27.12t

B:
X = 173.2 + 54.25cos(120)t + 0
X = 173.2 - 27.125t

x = x, so:
27.12t = 173.2 - 27.125t
t = 3.192921007 ≈ 3.19

A: X = 0 + 31.32cos(30)(3.19) ≈ 86.5
B: X = 173.2 + 54.25cos(120)(3.19) ≈ 86.7

Answer 8

answer is mot likely right; looks good :)

by habit and experience, i use -cos(60) instead of cos(120), but it's the same thing; force in the negative x direction.

sin(60) = sin(120), but cos(60) =/= cos(120) [ = negative cos(120)], so its always good to use the actual 120 to avoid having to remember and getting them confused.

glad i could help :)

Answer 9

So jealous of your knowledge! :]] Thank you!!