# What's the start speed and acceleration of the ball?

• Ockonal
In summary, the ball moves from the bottom to the top of the slant board, with a distance of 0.3 meters, after 1 and 2 seconds. To find the start speed and acceleration, the equation x = x_0 + V_0*t + (a*t^2)/2 is used, with initial position (x₀) and velocity (V₀) both equal to 0. By subtracting the two equations, V₀ can be expressed in terms of a. Substituting this into the first equation and solving for a, we get a value of -0.3 m/s^2. Using this value, V₀ can be calculated as 0.45 m/s. The acceleration
Ockonal

## Homework Statement

Using slant board, the ball was forced to move from bottom to the top. The ball was at the distance of 0.3 meters twice: after 1 and 2 seconds after start moving. What's the start speed and acceleration of the ball? (a=const)

## Homework Equations

$$\newline x = x_0 + V_0*t + \frac{a*t^2}{2}$$
x₀ = 0 m
V₀ = 0 m/s
l = 0.3 m
t₁ = 1 s
t₂ = 2 s

## The Attempt at a Solution

$$0.3 = 0 + V_0*1 + \frac{a*1^2}{2}$$

$$\newline 0.3 = 0 + V_0*2 + \frac{a*2^2}{2}$$
Here I can express, V₀. But I'm not sure the second expression is right. What's wrong?

Hi Ockonal, welcome to PF.
Now subtract first equation from the second equation. You will get Vo in terms of a.
Put it in the first equation to find a. Substitute this value in the second equation to get Vo.

rl.bhat said:
Hi Ockonal, welcome to PF.
Now subtract first equation from the second equation. You will get Vo in terms of a.
Put it in the first equation to find a. Substitute this value in the second equation to get Vo.
Oh, thanks.

Okay, I did the job:

$$0.3 = V_0 + \frac{a}{2}$$
$$0.3 = 2*V_0 + 2*a$$
=>
$$V_0 = 0.3 - \frac{a}{2}$$
$$0.3 = 2*(0.3 - \frac{a}{2}) + 2*a$$
$$0.3 = 0.6 + a;$$
$$a = -0.3 (m/s^2)$$
$$V_0 = 0.3 - \frac{a}{2};$$
$$V_0 = 0.15 (m/s)$$

The questions are:
Why the acceleration is negative? I think something is wrong in the first formulas.
The start speed is 0.45 m/s (refer to the answers), the acceleration is right (0.3)

The acceleration is negative because the ball is moving up the inclined plane.
Vo = 0.3 - a/2 = 0.3 - (- 0.3/2) = 0.3 + 0.15 = 0.45 m/s

Last edited:

## 1. What exactly is meant by "start speed" and "acceleration" of the ball?

"Start speed" refers to the initial velocity of the ball when it is first released or thrown, while "acceleration" refers to the rate at which the ball's velocity changes over time.

## 2. How is the start speed and acceleration of a ball measured?

The start speed and acceleration of a ball can be measured using various tools such as speed guns, radar guns, or motion sensors. These tools use different methods to calculate the velocity and acceleration of the ball.

## 3. Does the start speed and acceleration of a ball change depending on the type of ball?

Yes, the start speed and acceleration of a ball can vary depending on factors such as the mass, shape, and air resistance of the ball. For example, a heavier ball will typically have a lower start speed and acceleration compared to a lighter ball.

## 4. How do the start speed and acceleration of a ball affect its trajectory?

The start speed and acceleration of a ball play a significant role in determining its trajectory. A higher start speed and/or acceleration will result in a longer and more curved trajectory, while a lower start speed and/or acceleration will result in a shorter and less curved trajectory.

## 5. Is it possible to control or manipulate the start speed and acceleration of a ball?

Yes, the start speed and acceleration of a ball can be controlled and manipulated through various factors such as the angle and force of the initial throw or release, as well as the properties of the ball itself. For example, a smoother and more aerodynamic ball may have a higher start speed and acceleration compared to a rougher and less aerodynamic ball.

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