What Are the Minimum and Maximum Launch Speeds to Get a Match into a Basket?

[riddhish]

there is a waste paper basket of diameter D and height 2D. u throw a burning match from the same level as the bottom of the basket ,the horizontal distance between the near side of basket and the point from which u throw the match. u launch the match at 45 degrees with the horizontal. find the minimum and maximum values of the launch speed so that the match enters the basket.(ingore air resistance and give ans in terms of D and g).



please help 'sob'

 

[tiny-tim]

Hi riddhish!

First, weed out the irrelevant detail …

The question is the same as:
Launch an object at 45º so that, after it reaches its greatest height, it reaches height 2D between L and L + D, where L is the initial horizontal distance.

So start "Let the initial velocity be v", and carry on from there!

 

[Moetasim]

I would approach this problem by first breaking it down into smaller components and then using mathematical equations to solve for the minimum and maximum launch speed.

Firstly, we can consider the motion of the match as a projectile, which means it will follow a curved path due to the force of gravity. We can use the basic equations of projectile motion to solve for the horizontal distance between the near side of the basket and the point from which the match is thrown.

Using the given information, we know that the initial height (y0) of the match is 2D (same as the height of the basket) and the initial velocity (u) is unknown, but we can solve for it using the given angle of 45 degrees. The horizontal distance (x) can be calculated using the equation x = utcosθ, where θ is the angle of launch.

Now, to find the minimum and maximum values of the launch speed, we need to consider the trajectory of the match and ensure that it enters the basket. This means that the vertical distance the match travels (y) must be less than or equal to the radius of the basket (D/2).

We can use the equation y = y0 + utsinθ - 1/2gt^2 to solve for the time (t) it takes for the match to reach the height of the basket (y = D/2). We can then substitute this value of t into the equation for x to get an expression for the minimum and maximum launch speed.

Solving for u, we get u = (D/2) / (tcosθ). Substituting the value of t, we get u = (D/2) / (cosθ * √(2D/g)). This gives us the minimum and maximum launch speed in terms of D and g.

In conclusion, the minimum launch speed would be (D/2) / (cosθ * √(2D/g)) and the maximum launch speed would be (D/2) / (cosθ * √(2D/g)). However, it is important to note that in reality, air resistance would affect the motion of the match and these values may differ.