# Equation of Path of a Projectile

What path do projectiles follow? This can be seen by eliminating the time between the expressions for x and y.
....(ix)

Now, since g, Î¸o and are constants, Eq. (ix) is of the form y = ax+bx2, in which a and b are constants.
This is the equation of a parabola, i.e. the path of the projectile is a parabola.

Problem-1
A body is thrown horizontally from the top of a building. It falls on the ground after 1.5 s at a distance of 5 m from the building. Calculate (a) the height of the building (b) the horizontal velocity?

Solution
Time taken t = 1.5 s
Range x = 5 m
Acceleration due to gravity g = 9.8 ms-2
h =  gt2
h = Ã— 9.8 Ã— 1.5 Ã— 1.5
Height of the building h = 11.025 m
Range = horizontal velocity Ã— time taken
i. e., x = ut
Horizontal velocity u = = = 3.33 ms-1.

Problem-2

An aeroplane, flying horizontally at a height of 1960 m with a velocity of 125 ms-1 aims to hit an enemy target. Find at what distance from the tank, the pilot will release the bomb in order to hit the tank.

Solution

Acceleration due to gravity g = 9.8 ms-2

Height of the ground, h = 1960 m

Since the initial velocity is zero, h =  gt2

1960 =Ã— 9.8 Ã— t2

t2 = = 400

âˆ´ t = 20 s

Horizontal distance = Horizontal velocity Ã— time taken

= 125 Ã— 20 = 2500 m

The bomb should be released at a distance of 2500 m from the tank.

# Time of Maximum Height

What is the time taken by a projectile to reach the maximum height?

Let this time be denoted by tm. Since at this point, = 0,
= sin Î¸o - g tm = 0
Or, tm = sin Î¸o / g

The total time Tf during which the projectile is in flight can be obtained by putting y = 0 in Eq. (viii). We get
Tf = 2(sin Î¸o) / g

Tf is known as the time of flight of the projectile. We note that Tf = 2tm, which is expected because of the symmetry of the parabolic path.

# Maximum Height of a Projectile

The maximum height hm reached by the projectile can be calculated by substituting t = tm in Eq. (viii):
y = hm = (sin Î¸o)
Or, hm =

# Horizontal Range of a Projectile

The horizontal distance travelled by a projectile from its initial position (x=y=0) to the position where it passes y=0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight Tf. Therefore, the range R is
R = (cos Î¸o) (Tf)
= (cos Î¸o) (2sin Î¸o) / g
Or, R =         (x)

Equation (x) shows that for a given projection velocity , R is maximum if Î¸o = 45o since sin 90o = 1, the maximum value of sin 2Î¸o.

Thus the maximum horizontal range is
Rm =