# 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)

_{o}and are constants, Eq. (ix) is of the form y = ax+bx

^{2}, 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 = gt

^{2}

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 = gt^{2}

1960 =Ã— 9.8 Ã— t^{2}

t^{2} = = 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 t

_{m}. Since at this point, = 0,

= sin Î¸

_{o}- g t

_{m}= 0

Or, t

_{m}= sin Î¸

_{o}/ g

_{f}during which the projectile is in flight can be obtained by putting y = 0 in Eq. (viii). We get

T

_{f}= 2(sin Î¸

_{o}) / g

_{f}is known as the time of flight of the projectile. We note that T

_{f}= 2t

_{m}, which is expected because of the symmetry of the parabolic path.

# Maximum Height of a Projectile

The maximum height h_{m}reached by the projectile can be calculated by substituting t = t

_{m}in Eq. (viii):

y = h

_{m}= (sin Î¸

_{o})

Or, h

_{m}=

# 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 T_{f}. Therefore, the range R is

R = (cos Î¸

_{o}) (T

_{f})

= (cos Î¸

_{o}) (2sin Î¸

_{o}) / g

Or, R = (x)

_{o }= 45

^{o}since sin 90

^{o}= 1, the maximum value of sin 2Î¸

_{o}.

Thus the maximum horizontal range is

R

_{m}=