# Projectiles Revision

## Projectiles Revision

In these questions, take g to be 10m/s^{2}

### Question 1

A particle is projected with a velocity of 30 m/s at an angle of arctan(0.75) to the horizontal. It hits the ground at a point which is level to the point of projection. Find the time for which it is in air.

### Solution 1

*As long as a projectile lands at a point level to the point of projection, we can use the general projectile equations.*

This question requires the time of flight so we use the following equation:

### Question 2

A particle is projected with a velocity of 10m/s at an angle of 45° to the horizontal. It hits the ground at a point 3m below its point of projection. Find the time for which it is in air and the horizontal distance covered by the particle in this time.

### Solution 2

*Since the projectile is landing at a point that is not level to its point of projection, we cannot use the general projectile equations, so we use the general equations of motion.*

From what we can gather here: initial velocity (u) is 10m/s, the angle of projection is 45 ° and the final vertical displacement is -3m (i.e. 3m below the point of projection).

initial vertical velocity (u_{y}) = `u sin θ`

initial horizontal velocity (u_{x}) = `u cos θ`

With the data above, we can use the following equation to find time:

*Since the particle is thrown upwards which is against the force of gravity, the value of the acceleration due to gravity is negative.*

Now that it has reduced to a quadratic equation, we then solve the quadratic equation for the value of `t`.

Then we discard the negative value of `t`

Now that we found the time for which the particle is in the air, we can then use that time to find the horizontal distance covered as follows:

*Since in projectile we take air resistance to be negligible, there is no acceleration or deceleration in the horizontal direction, which means a = 0. This leaves the equation as:*

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