Trajectory formula is given by

$$y=xtan\theta -\frac{g{x}^2}{2v^{2}co{s}^2\theta }$$

Where,
y is the horizontal component,
x is the vertical component,
g= gravity value,
v= initial velocity,

= angle of inclination of the initial velocity from horizontal axis,

Trajectory related equations are:

$$TimeofFlight:t=\frac{2v_{0}sin\theta }{g}$$

$$Maximumheightreached:H=\frac{V_0^{2}si{n}^2\theta }{2g}$$

$$HorizontalRange:R=\frac{V_0^{2}sin2\theta }{g}$$

Where,
V_o is the initial Velocity,
sin

is the y-axis vertical component,
cos is the x-axis horizontal component.

Solved examples

Question: Marshall throws a ball at an angle of . If it moves at the rate of 6m/s and Steve catches it after 4s. Calculate the vertical distance covered by it.

Solution:

Given,

time = 4 sec

The horizontal distance is given by:

x = 24 m