**Projectile motion refers to the motion of an object that is projected into the air at an angle to the horizontal.**

Assumption:

- Acceleration due to gravity, g, is constant (9.81 $\text{m s}^{-2}$) over entire motion and is always directed downwards.
- No horizontal acceleration
- Effect of air resistance is negligible

**Path of projectile without air resistance (always a parabola) **

**Taking upwards as positive, (Click on image to enlarge)**

**Projectile motion with air resistance**

Key Characteristics of Graph:

- Horizontal range of object is shorter
- Maximum height reached is smaller
- Path is asymmetrical about highest point

Horizontally,

- Horizontal velocity no longer constant
- Horizontal velocity decreases with time. Object slows down
- Smaller horizontal range

Vertically,

- Going up, $a_{\text{upwards}} \, = \, g + \frac{F_{\text{Drag} \, \text{Force}}}{m}$
- Going down, $a_{\text{downwards}} \, = \, g-\frac{F_{\text{Drag} \, \text{Force}}}{m}$

Since $\mid a_{\text{upwards}} \mid \, > \, \mid a_{\text{downwards}} \mid$, $\Delta V_{\text{upwards}} \, > \, \Delta V_{\text{downwards}}$. So, object slows down at a faster rate on the way up, time taken to reach maximum point is less than the time taken to reach the ground from the maximum point.

Simulations of projectile motion: http://phet.colorado.edu/en/simulation/projectile-motion

I need to know what if the air resistance is not neglected Will The optimum angle be greater or less than 45 degrees

a canon ball is project so as to attain a maximum range. Find the maximum height attained if the initial velocity is U. pls can someone help with this question

for second picture, last column, isn’t displacement positive? since you are taking highest point as reference and upwards as positive.

Nope. Taking upwards as positive means:

– upwards displacement from reference point is positive,

– moving in the upwards direction is positive (positive velocity),

– accelerating in the upwards direction is positive.

Hope this clarifies.