# Projectile motion in physics, what is it?

## What is Projectile motion in physics

Projectile is an object thrown into air or launched in space to move freely by itself under the influence of gravity and air resistance.

In projectile  motion we have time of Flight, time to reach maximum height, Maximum height and Range, all these are the parameters we take into consideration when talking about Projectile

### Time of Flight (T) in projectile motion

The time of flight (T) in projectile motion is defined as the time taken by projectile to reach the maximum height and return. The time of flight is given as:

$T=\quad \frac { 2USin\theta }{ g }$

Where U= initial velocity, g = gravitational force and Ѳ=angle of projection

### Maximum height (H) in projectile motion

The maximum height in projectile motion is defined as the highest vertical distance reached as measured from horizontal projection plane.  Maximum height is given as:

$H=\frac { { U }^{ 2 }{ Sin }^{ 2 }{ \theta } }{ 2g }$

### The range (R ) in projectile motion

In a projectile motion, range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane. The range is given as :

$R=\frac { { U }^{ 2 }{ Sin2\theta } }{ g }$

### Evaluation in projectile motion

1. A tennis ball hit with a velocity of 3m/s at an angle of 60o to horizontal, take g =10m/s2. Calculate the:
• Time of flight
• Maximum height
• Range

# Answer

Solution for (a)

$Time\quad of\quad flight\quad T\quad =\frac { 2USin\theta }{ g } =\frac { 2\times 3Sin60 }{ 10 } =\frac { 2\times 3\times 0.866 }{ 10 } =0.52s$

For Maximum height

$Maxi.\quad height=\frac { { U }^{ 2 }{ Sin }^{ 2 }{ \theta } }{ 2g } =\frac { { 3 }^{ 2\times }{ Sin }^{ 2 }{ 60 } }{ 2\times 10 } =\frac { 9\times { 0.866 }^{ 2 } }{ 20 } =0.3375m$

For range

$Range=\frac { { U }^{ 2 }{ Sin2\theta } }{ g } =\frac { { 3 }^{ 3 }{ Sin120 } }{ 10 } =\frac { 9\times { 0.866 } }{ 10 } =0.779m$

1. A Cannon ball is projected so as to attain a maximum range. Find the maximum height attained if the initial velocity is U

# Answer

1. A rocket is fired at 60o to the horizontal with an initial speed of 200m/s, taking g =10m/s2. Calculate its time of flight and its range on horizontal plane

Solution

$Time\quad of\quad flight=\frac { 2USin\theta }{ g } =\frac { 2\times 200\times Sin60 }{ 10 } =\frac { 400\times 0.866 }{ 10 } =34.64s$

$Range\quad =\frac { { U }^{ 2 }{ Sin2\theta } }{ g } =\frac { { 200 }^{ 2 }{ Sin120 } }{ 10 } =\frac { 40000\times 0.866 }{ 10 } =3464m$

# Answer

1. An anti-aircraft gun fires at an elevation of 60o at an enemy aircraft at 10000m above the ground. At what speed must the cannon be shot to hit the plane at that height? Take g =9.8m/s2

Solution

$Max.\quad Height\quad H=\frac { { U }^{ 2 }{ Sin }^{ 2 }{ \theta } }{ 2g } \\ \\ \Rightarrow { U }^{ 2 }=\frac { 2gH }{ { Sin }^{ 2 }{ \theta } } =\frac { 2\times 10\times 10000 }{ { Sin }^{ 2 }{ 60 } } =\frac { 200000 }{ { 0.866 }^{ 2 } } =\frac { 200000 }{ 0.749956 } =266682.3\\ \\ U=\sqrt { 266682.3 } =516.4m/s$

1. Which of the following statement about range , time of flight and maximum height in projectile motion is correct?
• A range depend on the angle of projection only
• The time of flight is shorter , the higher the velocity of projection
• Maximum height is the minimum vertical height reached in a projectile motion
• Range is the maximum horizontal distance reached  by the projectile

# Answer: D

Questions and Answers on the followings:

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