Projectiles

Projectiles Revision

Projectiles Revision

In these questions, take g to be 10m/s2



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

Projectiles

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:

Projectile time of flight equation

projectile calculation

Time of flight calculation

Time of flight

Time of flight = 3.6s



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

Projectile trajectory

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 (uy) = u sin θ

initial horizontal velocity (ux) = u cos θ

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

Displacement equation

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

displacement equation

Displacement equation

Displacement equation

Quadratic equation projectile

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

Quadratic formula

Quadratic formulae

Quadratic formula

Answer

Then we discard the negative value of t

T = 1.76s

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:

Displacement

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:

Horizontal displacement

Horizontal displacement

S = 17.6 m



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