What Was the Initial Speed of a Ball Thrown Horizontally from a Building?

[Aquas]

Hey, everyone! Glad I found this forum, I plan to be using it a lot to try to expand my Knowledge, I currently taking an AP Physics class in my high school with no prior physics experience. The teacher however is really great! Anyway let's see what we got going here.
(Note ▲=Delta)

Homework Statement

A ball is thrown horizontally from the roof of a building 58 m tall and lands 50 m from the base. What was the ball's initial speed?

Vertical Known
▲Y=-58
A Sub Y = -9.8m/s
t=5.918s
VnotY=?

Horizontal Known
▲X=50m
VnotX= ?

Homework Equations

t=5.918
I got this because assuming that the distance of 58m was traveled going a speed of -9.8m/s (g) it equals 5.918! (Hopefully)

The Attempt at a Solution

Its a long shot but I attempted to compare it to a problem done in school which was essentially the same but, involved an angle of the stone coming off the roof. Also different Knows were given. So this was my go&delta Y=(VnotX)(t)² + (.5)(g)(t)² I put in what I know

-58=(VnotY)(5.91)² + (4.9)(5.91)² Attempted to solve for VnotY

-58=(VnotY)(34.9281) + (171.14)

(VnotY)=(34.9281)+(229.14) I added the -58 over

(VnotY)=264.07 : ( I don't think this is right...

So that is where I am, I may be completely off...

Ok! So I looked at the problem again found an Inherent flaw,

▲y=(VnotY)(t)+(.5)(g)(t)²

SO! It turns out (VnotY) is Zero...Making that all Null and void...
So with some Algebra Magic I get t=3.43! With this I hope to get further...Will keep posted!

Another Update!
While looking through my book found that while somthing is in free fall it has 0 Acceleration in the X-Axis! So that voids Half of the Equation! Woo Go Physics! Making it simple and I found that the initial speed was 14.57

 

[IITian]

well done!

 

[tomcue]

m/s!

Hello there! It's great to see that you are actively seeking to expand your knowledge and using forums like this to help you with your studies. Let's take a closer look at the problem you have presented.

The first step in solving any physics problem is to clearly identify what is given and what is asked for. In this case, you are given the height of the building (58 m) and the horizontal distance the ball travels (50 m). You are asked to find the initial speed of the ball.

Next, it's important to think about what principles and equations can be applied to solve this problem. One key principle to remember is that the acceleration of an object in free fall is always -9.8 m/s^2. This applies to the vertical motion of the ball in this problem.

Now, let's take a closer look at your attempt at a solution. It's great that you attempted to use the equation for vertical motion (▲y = v0y * t + (1/2) * g * t^2). However, there are a few errors in your calculations. First, the value for t that you have used (5.918 s) is incorrect. This value represents the total time of flight for the ball, not the time it takes for the ball to reach the ground. To find the time it takes for the ball to reach the ground, you can use the equation ▲y = v0y * t + (1/2) * g * t^2, where ▲y = -58 m (since the ball is starting at a height of 58 m) and v0y = 0 (since the initial vertical velocity is 0). Solving for t, you should get t = 3.43 s.

Next, you attempted to use the equation ▲y = v0y * t^2 + (1/2) * g * t^2, which is incorrect. The correct equation for vertical motion is ▲y = v0y * t + (1/2) * g * t^2. Also, when you substituted in your known values, you incorrectly used 4.9 as the value for g. Remember, g is -9.8 m/s^2. Using the correct values, you should get v0y = 14.57 m/s, which is the correct initial vertical velocity of the ball.

Now,