Projectile motion: Height and Time of flight
Concepts on height of projectile motion:
1. Let us consider the particle is projected with initial speed u at an angle 
with ground.
As the particle just clears two vertical wall each of height h at time t1 and t2 respectively then, h = usint1 – 
——— (i)
and h = usint2 – 
——- (ii)
Comparing equation (i) and (ii) we get, usint1 – 
= usint2 –
Or, usin
(t1 – t2) = 
Or, usin = 
——- (iii)
Substituting the value of usin in equation (i) we get, h = t1 – = 
.
2. A man is standing at the origin fires a gun towards a target and it hits the target after 4 second of firing. The position vector of the target is (40
+ 20
+ 80
) ms-1. If x and y axes are on the horizontal plane then what is the velocity of projection of the bullet? [Take g = 10 ms-2].
Let us consider the projected velocity of the bullet is 
= 
+ 
+ 
.
Along x axis, 40 = vx.4 [acceleration = gcos900 = 0] or, vx = 10
Along y axis, 20 = vy.4 [acceleration = gcos900 = 0] or, vy = 5
Along z axis, 80 = vz.4 – 
Or, vz = 
= 40
Therefore, projected velocity of the bullet is = (10 + 5 + 40) ms-1.
Concepts on time of flight of projectile motion:
1. If two particles are projected from a point and the maximum heights for two projectiles remain same then the time of flight for these two projectiles are also same.
A particle is projected at an angle 
with speed 
then the maximum height is H1 = 
and time period T1 = 
.
Another particle is projected at an angle 
with speed 
then the maximum height is H2 = 
and time period T2 = 
.
As, H1 = H2 so, =
Or, u1sin
1 = u2sin2
So, T2 = = = T1.
2. Determine the time of flight of a particle in between two points on a horizontal plane in projectile motion:
Let us consider, a particle is projected at an angle with horizontal (X axis) with initial speed u.
The time of flight of the particle during its projectile motion is T = 
.
If H is the maximum height reached by the particle during its motion then, H = 
.
Therefore, T2 = 
= (
)() = 
Let us consider A and C are the two points on the path of the projectile motion of the particle on a horizontal plane at a height h as shown in figure.
If t is the time required to move the particle from A to C then, t2 = 
Therefore, t = 
.
3. A ball is freely falling from a height H and collides the surface of an inclined plane at a height h from ground as shown in figure. The direction of velocity of the ball is horizontal after collision with the surface. Calculate the value of h, so that the ball will take maximum time to reach the ground.
During the free fall motion of the ball, it moves the distance (H – h) for time t1. So, H – h = 
or, t1 = 
The ball takes t2 time to reach the ground after travelling vertical distance h. So, h = 
or, t2 = 
.
Total time taken by the ball to travel height H is t = t1 + t2 = +
For maximum time, 
= 0
Or, 
[ +] = 0
Or, 
[ – 
+ 
] = 0
Or, H – h = h [as g ≠ 0]
∴ h = 
.
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