• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Trajectory Problem

  • Thread starter acpyrus
  • Start date
I am not getting the same answer as the text for this problem:
A cannon is fired from the top of the castle wall. The cannonball is fired at 50 m/s at an angle of 30 degrees. A cannonball that was accidentally dropped hits the moat below in 5s.
(a) How far from the castle wall does the cannonball hit the ground?
(b) What is the ball's maximum height above the ground?

Known: x(i) = t(i) = 0
v(i) = 50 m/s
v(ix) = (50m/s)(cos30) = 43m/s = v(fx)
v(iy) = (50m/s)(sin30) = 25m/s
a(x) = 0
a(y) = -9.8 m/s^2
y(f) = 0
For (a), I need to solve for x(f)

Step 1. Given t=1.5s for a cannonball to drop to the ground, I can calculate the height of the castle wall or y(i) = 1/2 (-9.8m/s^2) (1.5s)^2 = 11m

Step 2. Using the above value, solve for t:
y(f)=0=y(i) + v(iy)(t) + 1/2a(y)(t)^2
-11 = 25(t) - 4.9(t)^2
-11 = t[25 - 4.9t]
t = 0 or let 25 - 4.9t = 0
t = -25/-4.9 = 5.1s

Step 3, calculate x(f) = x(i) + v(ix)(t) + 1/2a(x)(t)^2
x(f) = 0 + 43(5.1) + 0 = 219 m

The answer in the text says it should be 239 m. Any insight on what I have done wrong?

How do I calculate (b)?

Thanks and sorry for the long post.
 

andrevdh

Homework Helper
2,126
116
I think the problem is where you are solving for t. You have -11 on the lefthand side of the equation not zero!
 
Answe of the (B)

When the ball reaches its maximum height above the ground, its velocity became zero, so: (assuming that g=10)
The acceleration in the y direction is constant and equal to “-g” so we can write:
v(fy)*v(fy) – v(iy)*v(iy)=-2gh and we know that v(fy)=0 so
h = v(iy)*v(iy)/2g=(25*25)/(2*10) = 31.25
H=h+h(i) which h(i) is the height of castle wall
 
How should I solve for t? I revised my Step 2 above and came up with t=11s or t=7.3s --> both answers put my x(f) value way over what the answer is supposed to be.
 

andrevdh

Homework Helper
2,126
116
The general solution of a quadratic formula of the form:
[tex]ax^2+bx+c=0[/tex]
is given by:
[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
It is derived by completing the square of the above equation. In this case we need to solve for the times when the equation [itex]-4.9t^2+25t+11[/itex] is zero, therefore
[tex]\begin{equation}\begin{split}a=-4.9 \\
b = 25 \\
c = 11
\end{split}\end{equation}[/tex]
in this case. Use the second equation above to solve for the times - I got 5.5 s
 
Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top