A projectile is something that is thrown from the ground (or some height) with certain velocity directed at an angle from the vertical. Since there is uniform gravitational acceleration **g** downwards, the projectile follows a curved path. We neglect air resistance and assume no horizontal acceleration.

Usually this velocity(u) and the angle it makes with vertical(θ) is given, and we have to calculate the height(H) and range(R) of the trajectory. We can do this by considering the motion in x(horizontal) and y(vertical) directions separately. Let’s see how it works out.

The velocity can be broken down into horizontal and vertical components as

** u** = u cos θ î + u sin θ ĵ

#### Vertical motion:

The projectile has upward (positive) velocity( **u**_{y} = u sin θ) and experiences downward(negative) acceleration due to gravity (**g**). At maximum height H, the upward velocity becomes zero.

We know the Kinematics equation

v^{2} = u^{2} +2×a×s

Hence, we can write for the height reached H and acceleration (a = -g) as

0 = (u sin θ)^{2} – 2×g×H

which gives us

H = (u sin θ)^{2}/2g

The time taken to reach this height can similarly be calculated from

v = u +at

with a = -g. Hence

0 = u sin θ – g×t

⇒ t = (u sin θ)/g

When the projectile reaches the height H, it starts descending again and reaches the ground with same vertical speed as it had started. By symmetry, the total time of flight then becomes twice of the time taken to reach height H.

Hence

T = 2×t =2(u sin θ)/g

#### Horizontal Motion:

In the horizontal direction, there is no acceleration. The total time taken during motion is T. Hence, horizontal distance covered aka Range would be

R = (u cos θ)×T = (u^{2}sin 2θ)/g