Is this even solvable? Kinematics.

[tophat]

Homework Statement

A ball is thrown vertically upward from ground level with an initial velocity V0. The ball rises to a height, h, then lands on the roof of a building of height 1/2h. The entire motion requires 10s. Find the height, h, and the initial velocity, V0.

Homework Equations

Kinematics.

The Attempt at a Solution

I can't come up with anything, I'm using a final velocity of 0 m/s when running everything through and none of the values are making sense. I just need to know if it's actually solvable for real values. I'm in a hurry for work right now, but I will check back later.

I'm using: t2: 10s
t0= 0s
v1= 0 m/s (turning point)
a = -g
y1= h
y2= 1/2h
v2= 0 m/s (stop)
y0 = 0 m

 

[jackarms]

I think using that as your final velocity is your main problem so far. Yes, the final velocity will be zero once the ball hits the building and stops, but you can't use that value because the process of the object stopping isn't free-fall motion. You have to use the value of the velocity just before it lands as the final velocity.

 

[SteamKing]

You've just jotted down some trivial equations so far; you haven't actually written the proper kinematic equations involving distance as a function of time.

Yes, there is a numerical solution for V0 and h, and there is a fair amount of algebra and logical deduction involved in obtaining it.

 

[Gzyousikai]

$$s(t)=v_0t-1/2gt^2, s(t)_{\text{max}}=h=f(v_0,g),$$ solve $$s(t)=\frac{h}{2}$$ for t, and then plugin t=10s, get the larger answer

 

[HalfLostAndSearching]

(ground level)



Yes, this problem is solvable using kinematics equations. However, it is important to first clarify some assumptions and clarify the problem statement.

First, it is important to note that the problem does not specify the value of acceleration due to gravity, g. This value can vary depending on the location on Earth, so it is important to clarify which value should be used. For simplicity, let's assume g = 9.8 m/s^2.

Next, the problem states that the ball is thrown vertically upward, but it does not specify the angle at which it is thrown. This means that the initial velocity, V0, can have both a vertical and horizontal component. Let's assume that the ball is thrown with a vertical velocity component only, meaning that the horizontal component is 0 m/s.

Now, let's use the kinematics equations to solve for the unknowns, h and V0. We can use the following equations:

1. v = u + at
2. s = ut + 1/2at^2
3. v^2 = u^2 + 2as

Using the first equation, we can solve for the initial velocity, V0:

0 = V0 + (-9.8)(10)
V0 = 98 m/s

Next, we can use the second equation to solve for the height, h:

h = (98)(10) + 1/2(-9.8)(10)^2
h = 490 m

Therefore, the height, h, is 490 m and the initial velocity, V0, is 98 m/s.

In conclusion, this problem is solvable using kinematics equations, but it is important to clarify some assumptions and carefully apply the equations to find the correct values.