# How to solve horizontally-launched projectile motion problems using equations of motion?

Projectile motion questions are common in A-level physics as they combine the idea of resolving vectors into their components and using the equations of motion ('suvat' equations) to calculate unknown quantities relating to an object's motion.

There are 2 types of typical projectile motion questions: horizontally launched projectiles (in which an object is launched horizontally from an elevated plane and follows a parabolic path to the ground) and non-horizontally launched projectiles (in which an obect is thrown in the air from the ground, rises to a peak height and then drops to the ground again). In this question, we deal with the case that the projectile is horiontally launched.

Before attempting to answer projectile motion questions, it is important to understand what a projectile is. A projectile is an object which is moving under its own inertia - in simpler terms, t*he only force acting upon it is gravity. *

For this example, the following 'suvat' equations are used:

Eqn 1) v = u + a*t

Eqn 2) s = ut+0.5*a*t^{2}

Eqn 3) s = 0.5*(u + v)*t

Eqn 4) v^{2} = u^{2} + 2*a*s

Eqn 5) s = v*t - 0.5*a*t^{2}

where s = displacement, u = initial velocity, v = final velocity, a = acceleration and t = time

**Question:**

**A ball is thrown from the top of a 10m high building with a horizontal speed of 5ms ^{-1}. Calculate the time it takes to drop to the floor and its horizontal displacement at the point when it reaches the floor.**

The first step in all projectile motion questions should be to understand that the velocity of the object being thrown, in this case the ball, has 2 components, its vertical component and its horizontal component. As a result, when writing out our 'suvat' equations, we must have separate equations for the horizontal and vertical components. So we begin by writing out the suvat lists for each component:

Horizontal:

S = ?

U = 5ms^{-1}

V = 5ms^{-1}

A = 0ms^{-2}

T = ?

Note that because gravity is the only force acting on the projectile and this is a vertical force, there is no horizontal force and hence no acceleration in the horizontal direction, so the intial and final velocities remain the same. So far, we don't know any other horizontal values.

Vertical:

S = -10m

U = 0ms^{-1}

V = ?

A = -9.8ms^{-2}

T = ?

In this case, the gravitational force creates an acceleration due to gravity equal to -g, where g = 9.8ms^{-2}. The inital vertical component of velocity is 0 as the object is thrown vertically (ie. there is no vertical component when t=0, the vertical component only occurs because of acceleration due to gravity as the object falls). Remember that displacement is a vector so the vertical displacement is the height with a minus sign in front of it to show that the ball has dropped this height.

The next step in this solution is to use the vertical suvat equations to calculate the time it takes the ball to reach the ground. This is done by looking at the suvat list and comparing to the equations above. In this case, we want to find t and we have u, a and s, so we use Eqn 2:

s = u*t + 0.5*a*t^{2}

Since u = 0, the equation simplifies to:

s = 0.5*a*t^{2}

Rearranging for t gives:

t = sqrt(2s/a)

**Subbing in s = -10 and a = -9.8 gives t = 1.429 seconds, this is the time it takes the ball to reach the floor.**

Now that we have calculated the time, we can use this is determine the horizontal displacement of the ball. Since time is the same in both suvat lists (the time taken for the ball to finish its path is the same whether you are looking at its horizontal or vertical velocity), we can add this t into the horizontal suvat list we wrote above:

Horiztonal:

S = ?

U = 5ms^{-1}

V = 5ms^{-1}

A = 0ms^{-2}

T = 1.429 s

So now we want to calculate s, and we have u, v and t, so we use Eqn 3:

s = 0.5*(u+v)*t

since u = v we can substitute this in to get:

s = 0.5*(2u)*t

s = u*t

so subbing in u = 5ms^{-1} and t = 1.429s, we can calculate **s = 7.145m. This is the horizontal displacement of the ball in the time it takes it to reach the ground.**