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

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SUMMARY

The discussion centers on calculating the horizontal distance a stone travels when thrown at a plum tree from a height. The stone is thrown at a speed of 35 m/s at an angle of approximately 16.26 degrees, with the plum located 3.675 meters above the throw point. The correct time of flight to reach this height is derived using the quadratic formula, yielding two solutions: t = 0.5s (ascending) and t = 1.5s (descending). The horizontal distance is calculated using the horizontal velocity of 33.6 m/s, leading to a final distance of 50.4 meters when the correct time is applied.

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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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