![projectile formula projectile formula](http://saxbyphysics.com/Images/ProjectileMotion.png)
#Projectile formula plus#
We will start with the following formula, which tells us that the displacement of an object undergoing constant acceleration is equal to its initial speed multiplied by time plus one-half of the acceleration it is experiencing multiplied by time squared. So we need an equation that relates displacement to the time during the projectile motion.
![projectile formula projectile formula](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/20161/content_an01.png)
To answer this question, we need to find the time at which the projectile reaches a certain displacement. The question asks us to work out the time at which it returns to the ground again, which we will call capital ?, and is also known as the time of flight of the projectile. And at the end of the motion, its vertical displacement is equal to zero. We will call the vertical displacement of the projectile ? ?. And we are told that the projectile returns to the ground at the same height that it was launched from. This downward force causes the projectile’s path to be curved. During projectile motion, there is only the downward force from gravity acting on the projectile, which has a magnitude of the mass of the projectile, which we will call ?, multiplied by ?, the acceleration due to gravity.
![projectile formula projectile formula](https://i.ytimg.com/vi/fcTTnpA-bgU/hqdefault.jpg)
We have a projectile launched with an initial speed that we will call ?, and it’s launched at an angle above the horizontal that we will call ?. What is the time between the projectile leaving the ground and returning to the ground at the same height that it was launched from? A projectile has an initial speed of 25 meters per second and is fired at an angle of 48 degrees above the horizontal.