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

Calculating height (projectile motion)

  • Thread starter roam
  • Start date
  • #1
1,266
11

Homework Statement



A rock is thrown from the edge of the top of a building with an initial velocity of 12.2 m/s at an angle of 53 degrees above the horizontal. The rock strikes the ground a horizontal distance of 25 m from the base of the building. Assume that the ground is level and that the side of the building is vertical. How tall is the building?

The correct answer is 23.5m.


The Attempt at a Solution



[tex]v_{ix}=12.2 cos 53 =7.34[/tex]
[tex]v_{iy}=12.2 sin 53 = 9.74[/tex]

[tex]t=\frac{v}{x}=\frac{7.34}{25}=0.29[/tex]

[tex]y=v_{iy}+1/2gt^2=0.29 \times 9.74 + \frac{1}{2}(9.81)(0.29)^2 = 3.23[/tex]

So why is my answer wrong?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Your formula for y is wrong. It should be
y = viy*t - 1/2*g*t^2.
 

Related Threads on Calculating height (projectile motion)

Replies
21
Views
3K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
5
Views
6K
  • Last Post
Replies
5
Views
10K
  • Last Post
Replies
1
Views
4K
Top