Answer:⬇️
Step-by-step explanation:
Assuming that the missile moves under constant acceleration due to gravity, we can use the kinematic equations to find the slant distance and time taken for the missile to reach point B.
First, we can find the initial velocity vector v0 of the missile by using the given horizontal and vertical components:
|v0| = √(100^2 + 45^2) ≈ 113.14 m/s
The angle θ that the initial velocity vector makes with the horizontal can be found as:
θ = arctan(45/100) ≈ 23.2 degrees
The time taken for the missile to reach point B can be found by using the vertical component of the velocity:
v_y = v0 sin(θ) = 45 m/s
Using the kinematic equation for vertical motion, we can find the time taken for the missile to reach the highest point of its trajectory:
v_y = v0 sin(θ) - gt
where g is the acceleration due to gravity, which is approximately 9.81 m/s^2. Solving for t, we get:
t = (v0 sin(θ))/g ≈ 4.60 s
Since the missile reaches the highest point of its trajectory halfway between points A and B, the total time taken for the missile to reach point B is:
2t ≈ 9.20 s
To find the slant distance d, we can use the horizontal component of the velocity:
v_x = v0 cos(θ) = 100 m/s
Using the kinematic equation for horizontal motion, we can find the slant distance d:
d = v_x t = (v0 cos(θ))(2t) ≈ 2068.8 m
Therefore, the slant distance from point A to point B is approximately 2068.8 meters, and the time taken for the missile to reach point B is approximately 9.20 seconds.