Acceleration and maximum range

[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..

 

[haruspex]

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

 

[bolskipolski]

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

I don't have any more info, so I don´t know

 

[haruspex]

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

 

[bolskipolski]

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

 

[haruspex]

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?

 

[bolskipolski]

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

 

[haruspex]

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

 

[bolskipolski]

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)

 

[haruspex]

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?

 

[bolskipolski]

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) ?

 

[haruspex]

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.