Two projectiles A and B are projected at the same time in the same vertical plane. A is projected at a height of 2m above the ground making an angle of 30deg with the horizontal. B is projected with the velocity of 20m/s at an angle of 60deg with the horizontal. If they collide. Determine;

i) initial velocity of A (found it=11.5m/s)
ii)horizontal distance moved on the point of collision
iii)the time taken when they collide.

Let’s take as the origin of the coordinate system the point where the projectile B is projected. The point of projectile A start is h = 2 m (above origin).

For A:
x(A) =v(oxA)•t =v(oA)•cosα•t,
y(A) =h +v(oyA)•t - g•t^2/2 =
=h +v(oA)•sinα•t - g•t^2/2.
For B:
x(B) =v(oxB)•t =v(oB) •cosβ•t,
y(B) = v(oyB)•t - g•t^2/2 =
=v(oB)•sin β•t - g•t^2/2.
For collision point:
x(A)= x(B), and y(A) =y(B), and the same “t”. Then
v(oA)•cosα•t = v(oB)•cosβ•t, =>
v(oA)=v(oB)•cosβ/ cosα =20•cos60/cos30 = 11.5 m/s.

h +v(oA)•sinα•t - g•t^2/2 =
=v(oB)•sin β•t - g•t^2/2.
t = h/{v(oB)• sin β - v(oA)•sinα} =
= 2/{20•sin 60 -11.5•sin30} = 0.17 s.
y(B) = v(oB)•sin β•t - g•t^2/2 = =20•sin60•0.17 =9/8•(0.17)^2/2 = 2.8 m.
x(B) =v(oB)•cosβ•t =20•cos60•0.17=1.7 m.
Coordinates of the collision point are (1.7 m, 2.8m).

thnakss soo much

I don't understand the solution of the question

To solve this problem, we can use the principles of projectile motion. Let's break down the problem step by step.

i) Finding the initial velocity of projectile A:
- Given that the height of A is 2m. We can use this information to find the vertical component of the initial velocity.
- The vertical component can be found using the formula: v₀ₐy = vₐ * sin(θₐ), where vₐ is the initial velocity of A and θₐ is the angle of projection.
- Since θₐ = 30° and we are given that vₐ = 11.5m/s, we can substitute these values into the formula:
v₀ₐy = 11.5 * sin(30°)
v₀ₐy = 5.75 m/s
- Now, we need to find the horizontal component of the initial velocity. We can use the formula: v₀ₐx = vₐ * cos(θₐ).
- Since θₐ = 30° and we are given that vₐ = 11.5m/s, we can substitute these values into the formula:
v₀ₐx = 11.5 * cos(30°)
v₀ₐx = 9.98 m/s (rounded to two decimal places)
- Therefore, the initial velocity of projectile A is approximately 9.98 m/s.

ii) Finding the horizontal distance moved at the point of collision:
- Since projectile B is projected at an angle of 60°, we can use its horizontal component of velocity to find the distance it traveled.
- We can use the formula: s = v₀ₐx * t, where s represents distance, v₀ₐx is the horizontal component of the initial velocity of A, and t is the time taken for them to collide.
- We need to find the time of collision, which leads us to the next step.

iii) Finding the time taken when they collide:
- To find the time of collision, we need to consider the time taken by both projectiles to reach the same vertical height.
- Since B is projected with a velocity of 20m/s at an angle of 60°, we can find its vertical component using the formula: v₀by = v₀b * sin(θb).
- We can substitute the values into the formula: v₀by = 20 * sin(60°) = 17.32 m/s.
- Now we can calculate the time taken by projectile B to reach the collision point using the formula: t = (2 * v₀by) / g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
- Substituting the values: t = (2 * 17.32) / 9.8 ≈ 3.53 seconds.
- Since both projectiles are launched at the same time, the time taken by projectile A to reach the collision point is also 3.53 seconds.

Now, we can use the time of collision (3.53 seconds) to calculate the horizontal distance moved at the point of collision using the formula: s = v₀ₐx * t.
Substituting the values:
s = (9.98 m/s) * (3.53 s) ≈ 35.16 meters.

Therefore:
i) The initial velocity of projectile A is approximately 9.98 m/s.
ii) The horizontal distance moved at the point of collision is approximately 35.16 meters.
iii) The time taken when they collide is approximately 3.53 seconds.