Question

A water balloon is launched at a speed of 24 m/s and an angle of 36 degrees above the horizontal. The water balloon hits a tall building located 26 m from the launch pad. At what height above the ground level will the water balloon hit the building? Calculate the answer in meters (m) and rounded to three significant figures

Answer 1

do you can drawing it please this case ? i think will be usefully

Answer 2

so far I have this drawn but I'm not sure if it's correct..

Answer 3

so this 26 m i think not will be this imagined by you there
what mean 26 m from the launch pad ?

Answer 4

Lol, kinematic equations and 2d problems. Physics I bet.

So your launch angle is 36 degrees, and velocity is 24 m/s. That is correct. Forget the 24cm part for now.

What you need to do first is to divide the 24m/s into x and y components. Do you know decomposition of vectors?

Answer 5

Basically dividing the 24m/s into x m/s and y m/s

Answer 6

And then after that, you have to find the time the water balloon reaches 24m in the x-direction using the x-direction velocity component.

Answer 7

which equation would I use for that?

Answer 8

So to divide the 24m/s into components.

X-direction:

cos(36) * 24

That gives you it's speed in the x-direction only. Get it?

Answer 9

oh! so then sin(36)*24 for y direction, right?

Answer 10

Yes. We'll worry about that later though.

Answer 11

I got 19.42 for the speed in x-direction

Answer 12

Alrighty.

So then once you have the x-velocity, and assuming there's no air-friction.

We use \[x = x_{0} + v_{0}t + \frac{ 1 }{ 2 }at ^{2}\]

You were taught this formula right?

Answer 13

yeah, I recognize it. is it okay if I work on it and then you check my answers?

Answer 14

suuure

Answer 15

I got t=3.96

Answer 16

@mhchen

Answer 17

Erm hold on.

so x = 0 + vt + 0
24 = (19.42)t
t = (24/19.42)
t = 1.24s

I got something different :\

Answer 18

a = 0 and initial position = 0

Answer 19

oh, there's no gravity?

Answer 20

that's only for y-direction.

Answer 21

oh okay

Answer 22

Okay so now that you've got t.

You can do the same equation for y:

\[y = y_{0} + v_{0}t + \frac{ 1 }{ 2 }at ^{2}\]

And this time you would put the gravity in.

Answer 23

and I'm using what I got for sin right? or am I still using the cos

Answer 24

you use sin this time. sin finds the y-component.

Answer 25

I got 9.95

Answer 26

yeah same here.

Answer 27

I'm in physics as well. I remember doing these problems like last week. It gets more complicated later on :(

Answer 28

oh gosh.. I hate it. I'm confused like half the time :(
so would this be the height then?

Answer 29

Yeah.

Answer 30

thanks for your help, I appreciate it!

Answer 31

np