Easy Question on Projectile Motion

  • Thread starter waters
  • Start date
  • #1
29
0
Okay, so I don't know what it is I'm doing wrong here.

Here is the question:
A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no obstacles in its way. It is fired at an angle 33° with respect to the horizontal and with an initial speed of 36 m/s. Air resistance is negligible in this situation.

The same projectile was then fired in the same way toward a wall that is a horizontal distance 106.2 m from where the projectile was fired.
What was the height of the projectile when it hit the wall?

t = vx/dx = 106.2 m / 36cos(33degrees) = about 3.5 s
Peak is at 2 s: vfy = viy + at = 0 = 36sin(33degrees) - 9.8t
3.5 s - 2 s = 1.5 s
y = vyt + (1/2)(ay(t^2)) = 0 + (.5*9.81*1.51747^2) = 11.3 m

I got 11.3 m, and it is NOT the correct answer
 

Answers and Replies

  • #2
Doc Al
Mentor
45,089
1,396
t = vx/dx = 106.2 m / 36cos(33degrees) = about 3.5 s
OK.
Peak is at 2 s: vfy = viy + at = 0 = 36sin(33degrees) - 9.8t
3.5 s - 2 s = 1.5 s
y = vyt + (1/2)(ay(t^2)) = 0 + (.5*9.81*1.51747^2) = 11.3 m
What you found (assuming you did the arithmetic OK) is the distance it fell from the peak. What you need is the height from the ground.

Why not just plug the time into the equation for y?
 
  • #3
29
0
Thanks for the reply.

I'm feel as if I did the arithmetic correctly, but that's only because I don't think I can find my mistakes. I just entered 18.9 m, and it was wrong. The peak I calculated was at 30.192 m, and it was right.

By plugging the time into the equation for y, you mean vx/dx, right?
 
  • #4
29
0
Hold on, that was the speed at the top. My bad. I'll try again.

I'm not too great at physics; it is seriously one of my weak spots.

Thanks for the help; I got the right answer. The maximum vertical distance is 19.594, so 19.594 -11.29 = 8.3 m, which is the correct answer. Sorry about the sig figs.
 
  • #5
Doc Al
Mentor
45,089
1,396
The peak I calculated was at 30.192 m, and it was right.
Why do think it was right? (How far will something fall in 2 seconds?)
By plugging the time into the equation for y, you mean vx/dx, right?
No, I meant plug it into the formula for y(t). What's that formula?

But there are several ways to solve this. Yours is just fine, but you're making an error.
 
  • #6
29
0
Oh, what I meant by that was substituting vx/dx for time into the equation that I used for y(t). Actually, it's exactly how I solved this problem.
 

Related Threads on Easy Question on Projectile Motion

  • Last Post
Replies
14
Views
11K
  • Last Post
Replies
1
Views
868
Replies
1
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
2K
Replies
5
Views
7K
  • Last Post
Replies
3
Views
1K
Top