Question

A particle is projected from a point a with a velocity of 21m/s at an angle of 53.1 degees to the horizontal. one second later another particle is projected at a point 0.3m below a with initial velocity of 31.5m/s at an angle of 36.9 degrees to the horizontal.

(i) prove that the particles collide
(ii) find the time of collision
(iii) find the distance of collision from point of projection

Answer 1

(i) Please see graph of the motion of the particles created with MS Excel and the calculations in the following sections

(ii) The time of collision is approximately 1.0 seconds

(iii) The common horizontal distance of point collision from the point of projection is approximately 25.2 meters

Step-by-step explanation:

The velocity with which the first projectile was fired, v₁ = 21 m/s

The angle to the horizontal the particle is launched = 53.1°

The time at which the other particle was launched = 1 second later

The location from which the other particle was projected = 0.3 m below the first particle

The initial velocity of the second particle = 31.5 m/s

The angle to the horizontal at which the second particle was projected, θ = 36.9°

(i) The height reached, by each of the particle is given as follows;

y = u·t - 1/2·g·t²

For the first projectile, we have;

y = 21·(t₁+1)×sin(53.1°) - 9.81·(t₁+1)²/2 + 0.3

For the second projectile, we have;

y= 31.5·(t₁)×sin(36.9°) - 4.905·(t₁)²

If the two projectiles collide, we get;

21·(t₁+1)×sin(53.1°) - 9.81·(t₁+1)²/2 + 0.3 = 31.5·(t₁)×sin(36.9°) - 4.905·(t₁)²

Using a graphing calculator for simplifying, we get;

-11.93·t₁ + 12.2 = 0

t₁ = 12.2/11.93 ≈ 1.02

Therefore, at time t₁ = 1.02 seconds, after the launch of the second particle, the two particle will be at the same vertical height

However, whereby at the time, t₂, the particles collide, the horizontal distance travelled, 'x', will be equal;

We have;

x = u·cos(θ)·t₁

For the first particle, we have;

x₁₁ = 21 × cos (53.1°) × (t₂ + 1)

For the second particle, we have;

x₂₂ = 31.5 × cos (36.9°) × t₂

At the point of collision, we have;

x₁ = x₂

∴ 21 × cos (53.1°) × (t₂ + 1) = 31.5 × cos (36.9°) × t₂

31.5 × cos (36.9°) × t₂ - 21 × cos (53.1°) × t₂ = 21 × cos (53.1°)

t₂ = 21 × cos (53.1°)/(31.5 × cos (36.9°) - 21 × cos (53.1°) ) = 1.00219236871

t₂ ≈ 1.0 seconds

Given that t₁ ≈ t₂, the particles reach the same height and the same horizontal distance at the same time, t₂ ≈ 1.0 and therefore, they collide.

(ii) The time of collision is found above as t₁ ≈ t₂ ≈ 1.0 seconds

(iii) The horizontal distance of the point of collision from the starting point, 'x', is given as follows;

x = 21 × cos (53.1°) × (1.0 + 1) ≈ 25.2

The horizontal distance of the point of collision from the starting point, x ≈ 25.2 meters

The vertical distance of the point of collision from the starting point of the second particle, 'y', is given as follows;

y = 21 × (1+1)×sin(53.1°) - 9.81 × (1+1)²/2 + 0.3 ≈ 14

The vertical distance of the point of collision from the starting point of the second particle, y ≈ 14 meters

The magnitude of the distance from the starting point of the second particle, r = √(25.2² + 14²) ≈ 28.8

The magnitude of the distance from the starting point of the second particle, r ≈ 28.4 meters.