Question

a particle is projected at angle theta wirh horizontal from ground . the slope (m) of trajectory of particle varies with time (t) as

Answer 1

The solution is in that images.


The pages are numbered 1 and 2, at the top.


We are given a projectile motion.

By considering a Cartesian Co-ordinate System, we can analyse motion individually in X and Y Directions.

And by applying equations of motion, we can find both x and y coordinates as a function of time.


And then we can find the slope, as shown in the solution.

Now, we need the variation of slope with respect to time.

We can rewrite the equation as follows:

$\displaystyle m=\frac{u\sin\theta-gt}{u\cos\theta}\\\\\\\implies m=\frac{u\sin\theta}{u\cos\theta}-\frac{g}{u\cos\theta}t\\\\\\\implies m=\tan\theta-\frac{g}{u\cos\theta}t\\\\\\\implies \boxed{m=a-bt} \\ \\ \\ \implies \text{where }a=\tan\theta \quad and \quad b=\frac{g}{u\cos\theta}$

Since time is always positive, we get another condition that t>0.

If we represent the slope (m) on Y-axis and the time (t) on X-axis, then we get a representative graph of 

$y=a-bx$

with x>0

A sample graph is attached with sample values of a and b.

The graph represents qualitatively the variation of a slope of the trajectory of the particle with respect to time.

Answer 2

hope this will help you all