Find Initial Velocity - Projectile Motion

[Calaereth]

I've been trying to come up with the answer but it hasn't gotten me anywhere.
I know that I need to find the angle to get the velocity.

Q: A rock is projected from the edge of a 30.48 m tall building at some unknown angle above the horizontal. 5 seconds after, the rock strikes the ground at a distance of 48.76 m from the building. Determine the speed of the rock.

Given:
delta X = 48.76 m
delta Y = -30.48 m
t = 5 s
Find:
Vi
Angle (Not necessary)

 

[axmls]

What did you try? It helps to see where you're making mistakes in your work.

 

[Calaereth]

My teacher hasn't really taught us how to do it but gave us 5 motion equations:

a = V2 - V1 / delta t
delta d = 1/2 (V1 + V2) delta t
delta d =V1*t + 0.5t^2
delta d = v2*t - 0.5*a*t^2
v2^2 = V1^2 + 2a*delta d

I tried to find the angle using
theta = tan-1(Vyf/Vxf)
or
theta = tan-1(dy/dx)
It gave me 32 degrees

Tell me if this is right and where do I go from here?

 

[axmls]

How did you get the velocities for that angle formula?

 

[Calaereth]

I didn't since I used the distances:

theta = tan-1 (30.48m/48.76m)
=32 degree

 

[axmls]

But that's not telling you what angle it's thrown at--it's telling you the angle between the horizontal and the landing point.

 

[Calaereth]

Previously -
Horizontal
Vix = V cos 32
Vertical
Viy = V sin 32

I tried this:
delta Y = Viy*t +0.5 * -9.8 * 5^2
Viy = (delta y - 0.5*9.8*-t^2)/t
Viy = (30.48 -122.5)/5
Viy = -18.404 m/s

 

[Ahmad Hossain]

It is very basic question in physics. If you know motion Graph,I think it is very easy.You do not need to know the angel of the rock to find out its time to fall on ground.your method is also wrong.you should draw the diagram.

 

[Ahmad Hossain]

I hope you can solve this problem. If you cannot, than I will help you.

 

[Calaereth]

So basically I have to use simple equations to solve for this such as

V= d/t
V = 48.76 m / 5s
V = 9.75 m/s

If this is not right, then can you please help me?

 

[Ahmad Hossain]

it is also wrong

 

[Calaereth]

I've never had a problem that is to find the initial velocity of a projectile motion and with a height at that.

Let me try it out though.

 

[Ahmad Hossain]

what you study and which level you have? do not mind because it is easy for me to explain you.

 

[Calaereth]

In grade 12 and I don't have any previous knowledge on projectile motion. We haven't really covered it much in class.

 

[Ahmad Hossain]

yes, you read the book ...Physics for scientist and Engineers by tipler...Hope you will get the answer.

 

[Calaereth]

I need a faster approach to solving the problem. If you know how to solve it can you show me how?
If not then, I guess I can get the book.

 

[axmls]

So basically I have to use simple equations to solve for this such as

V= d/t
V = 48.76 m / 5s
V = 9.75 m/s

If this is not right, then can you please help me?

Yes, that is your horizontal velocity. Notice that the total horizontal distance is 48.76 m, and it is traveled in 5 s. You can do this because the x and y velocities are completely independent of each other.

Now you need to find the y component of the velocity and use both of them to find the total speed.

The formula you used before is correct, but you solved incorrectly. You have the wrong sign for the 0.5*9.8 t^2 term on the left side. Acceleration is negative, so you should be adding it to the left side. It should be $-30.48 + \frac{1}{2} 9.8 (5^2)$. Please do go through each step one by one to see that this is the case.

 

[axmls]

<Text removed by a mentor>

Yes, but please let Calaereth attempt it as they are the one who needs help.

 

[Calaereth]

Thank you very much. This has helped me a lot. :)