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

Projectile motion with differing heights

  • Thread starter Hedgehog1
  • Start date
  • #1
4
0

Homework Statement


A cannon is fired at a target on a cliff that is 400 meters away. The cliff has an elevation of 150 meters relative to the cannon. The initial velocity of the cannonball is 100 m/s. Ignoring wind resistance calculate the two possible angles and their respective times.
X= 400m
Y=150m
V_naught=100m/s
A= -9.81 m/s^2

Homework Equations



Time X = Time Y

X=V_naught*cos(theta)*t
Y= V_naught*sin(theta)*t - 0.5*a(t)^2


1/(cos(theta)0^2=1+ (tan(theta))^2



The Attempt at a Solution


t= 400/100cos(theta)

0= -0.5*a(t)^2 + V_naught*sin(theta)*t -150

0= -0.5*a(400/100cos(theta))^2 + V_naught*sin(theta)*400/100cos(theta) -150

I get stuck here. I'm not really quite sure what to do next. I'm pretty sure I have to use the quadratic equation, but don't know how.
 

Answers and Replies

  • #2
ehild
Homework Helper
15,396
1,803
Hi Hedgehog:smile:, welcome to PF.

Substitute vnought=100 into your equation and simplify. You also know that a=g, the gravitational acceleration.

ehild
 
  • #3
4
0
Thanks for the quick reply!

I substituted all my v0=100 and simplified to get this:

-0.5*-9.81(16/1+(tan(theta))^2) + 400*tan(theta) -150=0

Edit:
I'm not sure entirely how I would simplify further.
 
  • #4
ehild
Homework Helper
15,396
1,803
Your equation Y= V0*sin(theta)*t - 0.5*a(t)^2 is a bit incorrect, it has to be Y= V0*sin(theta)*t + 0.5*a(t)^2, with a=-9.81 m/s^2.
Using the parentheses correctly, you get
-0.5*9.81*16(1/cos2(θ))+ 400*tan(θ) -150=0.
You wrote among the relevant equations that 1/cos2(θ)=1+tan2(θ). Use it, substitute 1+tan2(θ)for 1/cos2(θ). You get a quadratic equation for tan(θ).

ehild
 
  • #5
4
0
Thank you for your help! I understand what to do now.
 
  • #6
ehild
Homework Helper
15,396
1,803
Thank you for your help! I understand what to do now.
You are welcome:smile:

ehild
 

Related Threads for: Projectile motion with differing heights

  • Last Post
Replies
0
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
10
Views
3K
Replies
7
Views
250
Replies
7
Views
15K
  • Last Post
Replies
8
Views
3K
Top