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

AI Thread Summary
Sandy throws a stone at a plum tree from point O at a speed of 35 m/s and an angle where cos α = 0.96. The stone's height above O is given by the equation 3.675 = 9.8t - 4.9t², leading to a quadratic solution for time t. The correct values for t are approximately 0.5 seconds and 1.5 seconds, with the latter being the relevant time for when the stone descends to the height of the plum. The horizontal velocity is calculated as 33.6 m/s, resulting in a horizontal distance of about 78.05 meters. Careless mistakes in calculations can significantly impact the final result.
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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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