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

In summary: But don't worry, we all make mistakes sometimes! Just keep practicing and you'll get even better at summarizing and solving problems.
  • #1
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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 [tex]\alpha[/tex] to the horizontal, where cos [tex]\alpha[/tex] = 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 [tex]\alpha[/tex] = 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:
 

Attachments

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  • #2
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.
 
  • #3
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 [tex]\alpha[/tex] = 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:
  • #4
Damn! Thanks guys. I feel like such an idiot for a careless mistake, costing me 6 marks!
 
  • #5
Anytime! And yeah, those little mistakes are always frustrating, I know the feeling.
 

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

1. What is a projectile in mechanics?

A projectile in mechanics refers to any object that is propelled into the air and moves under the influence of gravity. It follows a curved path, known as a parabolic path, due to the combined effects of its initial velocity and the force of gravity.

2. What are the key factors that affect a projectile's motion?

The key factors that affect a projectile's motion are its initial velocity, the angle at which it is launched, and the force of gravity. Other factors such as air resistance and wind can also have an impact on the projectile's path, but they are generally considered to be negligible in basic projectile problems.

3. How do you calculate the range of a projectile?

The range of a projectile is the horizontal distance it travels before it hits the ground. To calculate the range, you can use the formula R = (v02sin(2θ))/g, where R is the range, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

4. What is the maximum height of a projectile?

The maximum height of a projectile is the highest point it reaches in its trajectory. To calculate the maximum height, you can use the formula H = (v02sin2(θ))/2g, where H is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

5. How does air resistance affect a projectile's motion?

Air resistance, also known as drag, can have a significant impact on a projectile's motion. It acts in the opposite direction of the projectile's motion, slowing it down and causing it to deviate from its expected trajectory. In basic projectile problems, air resistance is often ignored, but it becomes more important for objects with larger surface areas or at higher velocities.

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