How Far Does the Stone Travel Horizontally to Hit the Plum?

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Homework Help Overview

The problem involves a stone thrown at a plum tree from a certain height and angle, with the goal of determining the horizontal distance to the plum. The context includes projectile motion, specifically focusing on the vertical height reached by the stone and the time it takes to reach that height.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the quadratic formula to find the time at which the stone reaches a height of 3.675m. There is mention of the need to consider two possible times due to the parabolic trajectory of the stone.

Discussion Status

Some participants have pointed out potential errors in the calculations presented, particularly regarding the quadratic solution. There is acknowledgment of the need to differentiate between the ascending and descending times when the stone reaches the specified height. The conversation reflects a collaborative effort to clarify misunderstandings and correct mistakes.

Contextual Notes

Participants are working under the assumption that air resistance is negligible and are focused on the mathematical interpretation of the projectile motion equations provided in the problem statement.

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Homework Statement



Sandy is throwing a stone at a plum tree. The stone is thrown from a point O at a speed of 35 ms^-1 at an angle \alpha to the horizontal, where cos \alpha = 0.96. You are given that, t seconds after being thrown, the stone is (9.8t - 4.9t^2)m higher than O.

When descending, the stone hits a plum which is 3.675m higher than O. Air resistance should be neglected.

Calculate the horizontal distance of the plum from O.

Homework Equations



Quadratic formula.
s = u t?


The Attempt at a Solution



alpha is 16.26 using arccos. using s=ut+1/2at^2 equation given, I use, 3.675 = 9.8t - 4.9t^2.
Then I use the quadratic formula (-b+-sqrt...ect) which comes out with t as 2.327s approx.
(When I did this, didn't think this was correct as the path is parabolic, so there are two points of the height 3.675m above O).

I find the horizontal vel. using 35cos \alpha = 33.6. so, using s=ut, i do 33.6x2.327... = 78.05.

This is all wrong apparently according to the mark scheme (attached). I don't understand the mark scheme.

Help! Thanks!:cry:
 

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It seems that you've solved the quadratic wrong. When I did it, I got different values. And yes, you should have two values when you solve the quadratic, the smaller of the two will be the time when the stone passes the 3.675m mark while ascending, and the larger one is when it is descending, which is the one that should be used.
 
MacDougal87 said:
alpha is 16.26 using arccos. using s=ut+1/2at^2 equation given, I use, 3.675 = 9.8t - 4.9t^2.
Then I use the quadratic formula (-b+-sqrt...ect) which comes out with t as 2.327s approx.
(When I did this, didn't think this was correct as the path is parabolic, so there are two points of the height 3.675m above O).

I find the horizontal vel. using 35cos \alpha = 33.6. so, using s=ut, i do 33.6x2.327... = 78.05.

Your calculation got problem because the answers should be :

-3.675+9.8t - 4.9t2=0 => t=0.5s / t=1.5s

:smile:
 
Last edited:
Damn! Thanks guys. I feel like such an idiot for a careless mistake, costing me 6 marks!
 
Anytime! And yeah, those little mistakes are always frustrating, I know the feeling.
 

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