# Acceleration and maximum range

• bolskipolski
In summary, we are trying to calculate the acceleration and maximum range of a jump for a frog. The given information includes a maximum vertical height of 0.28m and a distance of acceleration of 0.086m. The gravity constant is 9.81 m/s^2. The task is not entirely clear, but assuming the given height is for a max range jump and a take-off angle of 45 degrees, we can find the height and range by using kinematic equations. We also need to determine the take-off speed, which can then be used to calculate the duration of the jump and acceleration during take-off.
bolskipolski
How do you calculate the acceleration and the maximum range of a jump (in this case a frog,) when the maximum vertical hight is 0.28m and the distance of acceleration is 0.086 m? I also tried to use the 45 degree, and the gravitation constant 9,81 m/s^2 but can't get it right..

Is the max vertical height that's given for a vertical jump? For a 45 degree jump? Something else?

haruspex said:
Is the max vertical height that's given for a vertical jump? For a 45 degree jump? Something else?

Pls post the question as posed, word for word. You might be missing something in the interpretation.

haruspex said:
Pls post the question as posed, word for word. You might be missing something in the interpretation.

Task 2: Calculate the acceleration and maximum range of a jump in the case of a frog. The maximum vertical hight is 0.28m and the distance of acceleration is 0.086m. The gravity constant g=9.81 m/s^2

This is the task i got (word for word), nothing less nothing more

Ok, I read that as meaning the height given is for a jump of max range. (It's not entirely clear, but I think that's a fair reading.)
Suppose the take-off speed is u. What would the max range be? What height would be reached?

haruspex said:
Ok, I read that as meaning the height given is for a jump of max range. (It's not entirely clear, but I think that's a fair reading.)
Suppose the take-off speed is u. What would the max range be? What height would be reached?

could be, yes. my teacher don´t speak english, so he translates all the tasks in google translate ... Honestly, i have no clue what he is asking for.
So, i don't know, i don't know i don't know, and i am sorry for not answering your questions.
I just need a suggestion

You mentioned a 45 degree take-off. What made you suggest that?

haruspex said:
You mentioned a 45 degree take-off. What made you suggest that?

I read in a physicbook that the maximum distance of movement of an object in 2 dimensions, irrespective of wind resistance, the angle will always be equal to 45
(as evidenced by sin2 * 45 = 1)

On a level surface, yes. (It's different going up or down an incline.)
So assuming 45 degrees, and a take-off speed of u, what would the height and range be?

haruspex said:
On a level surface, yes. (It's different going up or down an incline.)
So assuming 45 degrees, and a take-off speed of u, what would the height and range be?

I don't have any take-off speed. The height i guess would be 0.28, the range could be the distance (0.086) ?

bolskipolski said:
I don't have any take-off speed.
I know that. I wrote, let the take-off speed be u, an unknown. If the take-off angle is 45 degrees, what is the vertical component of speed at take-off (as a function of u)? What kinematic equations do you have relating distance, constant acceleration and initial and final speeds? Using that, what height would be reached (as a function of u)? Knowing the height the frog does reach, what value does that give you for u?
the range could be the distance (0.086) ?
No, that's the distance through which the frog was accelerating during its take-off. After you've worked out the take-off speed, you can use it in the following ways:
- find out how long the frog takes to land, and thus how far he jumps
- find out what the acceleration must have been during the take-off period.

## What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning it has both magnitude and direction.

## How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. The formula for acceleration is a = (v2 - v1) / t, where a is acceleration, v2 is the final velocity, v1 is the initial velocity, and t is the time interval.

## What is maximum range?

Maximum range is the farthest distance an object can travel in a given direction. In physics, it is often used to refer to the maximum distance a projectile can travel before hitting the ground.

## How is maximum range affected by acceleration?

The greater the acceleration, the greater the maximum range. This is because a higher acceleration allows the object to cover more distance in the same amount of time, resulting in a longer range.

## What factors can affect acceleration and maximum range?

Factors that can affect acceleration and maximum range include air resistance, the angle of launch, and initial velocity. Other factors such as mass, friction, and external forces can also impact these quantities.

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