# How to solve a velocity problem as an initial condition problem

• Wolvenmoon
In summary, a small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 30.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Wolvenmoon

## Homework Statement

A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 30.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.

## Homework Equations

Acceleration of gravity = 9.81m/s^2

The normal way:
v^2 = u^2 + 2as
v=u+a*t

How I want to do it:
dv/dt = a(t)
ds/dt = v(t)

## The Attempt at a Solution

a(t) = -9.8 m/s^2
v(t) = -9.8t m/s + C(m/s)
17 = -9.8*0 + C, C=17
v(t) = -9.8t(m/s) + 17(m/s)
s(t) = -4.9t^2 + 17t + C
30 = -4.9*0 + 17*0 + C
C = 30
s(t) = -4.9t^2+17t+30
-30 = -4.9t^2+17t+30

t=5.64s for part B.
For part A, this would make it -9.8*5.64 + 17 = -38.272.

v=(17^2+2*9.81*30)^(1/2)=29.6m/s and 29.62=-17+9.81*t, t=4.76s

I'm certain there's a way to solve this as initial value problems. I'd rather get my calculus on than try to memorize these equations!

Wolvenmoon said:
s(t) = -4.9t^2+17t+30
-30 = -4.9t^2+17t+30

Why -30 on the left hand side?

I believe a bad guess. The initial position is y=0m (the level of the roof), the ball ends up at y=-30.

That probably should be a zero.

You need to decide. Either s(0) = 30, which fixes one value for the integration constant. Or s(0) = 0, which fixes some other value. Then you should have s(T) = 0 or S(T) = -30, respectively. But s(0) = 30, which you chose originally, is not compatible with with s(T) = -30 (because that would mean the street is 60 meters below the 30 meter tall building, which is impossible).

1 person
Thank you! That does provide the correct answer in either case (t=4.76, then that can be substituted into v(t) = -29.6m/s!)

I'm glad to know it was something simple.

## 1. How do I determine the initial velocity in a velocity problem?

The initial velocity can be found by analyzing the given information in the problem. If the problem provides the distance, time, and acceleration, you can use the formula v = u + at to solve for the initial velocity. If the problem does not provide this information, you may need to use other equations such as v = u + 2as or v^2 = u^2 + 2as to find the initial velocity.

## 2. What is the difference between initial velocity and final velocity in a velocity problem?

The initial velocity is the velocity at the starting point of an object's motion, while the final velocity is the velocity at the ending point. The difference between the two is that the initial velocity is used as a starting point to calculate the final velocity, and it is usually known or given in a problem. On the other hand, the final velocity is what you are trying to find in the problem.

## 3. Can I use an initial velocity problem to find the acceleration?

Yes, you can use an initial velocity problem to find the acceleration. If you are given the initial velocity, final velocity, and time, you can use the formula a = (v - u)/t to solve for the acceleration. However, if you are given distance and time instead, you will need to use other equations such as a = 2d/t^2 or a = (v^2 - u^2)/2d to find the acceleration.

## 4. How do I solve a velocity problem with non-constant acceleration?

If the acceleration in a problem is not constant, you will need to use calculus to solve it. You can use the formula v = u + at for each small time interval to find the average velocity, and then take the limit as the time interval approaches zero to find the instantaneous velocity. Alternatively, you can use the kinematic equations for non-constant acceleration, such as v = u + (1/2)(a + a0)t, where a0 is the initial acceleration.

## 5. What are some common mistakes to avoid when solving initial velocity problems?

Some common mistakes to avoid when solving initial velocity problems include using the wrong units, not considering the direction of motion, and not using the appropriate equation for the given information. It is also important to double-check your calculations and make sure they are consistent with the given information. Additionally, be careful not to round off values too early in the calculation, as this can lead to inaccurate results.

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