Question

discuss the horizontal projection of a projectile from a tower of height H find equation of trajectory time of flight and maximum horizontal range​

Answer 1

Explanation:

Projectiles

This secton covers projectiles revision.

When a particle is projected from the ground it will follow a curved path, before hitting the ground. How far the particle travels will depend on the speed of projection and the angle of projection.

The suvat equations can be adapted to solve problems involving projectiles.

Range

The range (R) of the projectile is the horizontal distance it travels during the motion.

Now, s = ut + ½ at2

Using this equation vertically, we have that a = -g (the acceleration due to gravity) and the initial velocity in the vertical direction is usina (by resolving). Hence:

y = utsina - ½ gt2 (1)

Using the equation horizontally:

x = utcosa (2)

Remember, there is no acceleration horizontally so a = 0 here.

When the particle returns to the ground, y = 0. Substituting this into (1):

0 = utsina - ½ gt2

t (usina - ½ gt) = 0

t = 0 or t = 2u sina (3)

g

Therefore when x = R, t = 2u sina / g

Substituting into (2):

R = ucosa (2u sina)

g

The range is therefore

Time of Flight

The time the ball is in the air is given by (3).

Time of flight =

Maximum Range

If a particle is projected at fixed speed, it will travel the furthest horizontal distance if it is projected at an angle of 45° to the horizontal. This is because the maximum sin2a can be is 1 and sin2a = 1 when a = 45°.

Velocity

The velocity of the particle at any time can be calculated from the equation v = u + at.

By applying this equation horizontally, we find that:

dx/dt = ucosa

By applying it vertically, we find that:

dy/dt = usina - gt

Hope this helps mate ☺️.