Newton's 2nd Law. Help needed urgent!!!!!!!!!!!!!?

Question

A crate of relief supplies and attached parachute which has a combined mass of 120 kg is dropped from an aircraft horizontally at 70ms^-1 at a height of 100m above the ground.
The crate falls for 2 seconds before the parachute opens automatically and instantaneously

In the freefall part of motion, the origin of coordinates is at the point on the ground vertically below where the crate was released from the aircraft. Model the crate/parachute as a
particle. Ignore air resistance.

1 what are the initial position and velocity of the crate

2 write down the equation of motion of the crate
Derive expressions for the velocity v(t) and position r(t)
of the crate (0 < t < 2) (v and r should be bold)

3 Calculate the speed and direction of motion of the crate at
the instant the parachute opens.

Answer 1

I couldn't tell from the problem statement which coordinate system was preferred, so we'll work with Cartesian coordinates (e.g., x and y) which is the easiest. You can always convert to radial coordinate (radius and angle) using the procedure given at the end of my reply.

1. These are given by the problem. The initial position is
r(0) = (0, 100),
where the first number is the horizontal coordinate and the second number is the vertical coordinate. The initial velocity must be
v(0) = (70, 0)
since the horizontal velocity is a constant 70 m/s and the velocity of a dropped object at time t = 0 is zero.

2. The velocity in the horizontal direction is constant 70 m/s, ignoring air friction. The velocity in the vertical direction increases with time since it is in freefall. The equation for the velocity of an object under gravitational acceleration is v = g t, where g = 9.8 m/s². So we can write an expression for the velocity vector as a function of time, v(t) = (70, - gt). You can derive an equation for the position by either integrating the equation for v(t) with respect to time, or by using the well-known equation for distance,
d = d0 + v0 t + ½ g t², where d0 = initial position, v0 = initial velocity. The position vector would then be
r(t) = (70t, 100 - ½ g t²)

3. You can calculate this by plugging in t = 2 into the above equation for velocity and solving. The speed would be the magnitude of the velocity vector, which is the square root of the sum of the squares of the horizontal and vertical velocity components. The direction can be calculated from taking the arc-tangent of the ratio of the two.

For example, at t = 2, v(2) = (70, -19.6). The magnitude would be the square root of 70² + 19.6² which is 72.7 m/s, and the angle from the horizontal would be arctan (-19.6/70) = -15.6 deg.

Answer 2

We’ll be working in both rectangular (x,y) and polar r∠ϴº coordinates. Where
r = √(x^2 + y^2) and ϴ = arctan(y/x) I use my calculator to make conversions.
g = accelaration of gravity = 9.8 m/sec^2
(I use r'(t) = v(t) below to remind me that it is the differential wrt time of the position)

1. from problem definition
Y0 = initial height = 100m x
X0 = initial X position = 0 m
R0 = initial combined position (X0 Y0) in polar coordinates = 100∠90º meters

Y’0 = initial vertical velocity = 0 m/s
X’0 = initial x velocity = 70 m/s
R’0 = initial combined velocity = (70 0) = 70∠0º m/s

2. x(t) = x’0*t + X0 = 70t m
y(t) = Y0 – gt^2/2 = 100 – 4.9t^2 m
r(t) = √(x(t)^2 + y(t)^2) ∠ arctan(y(t)/x(t)) m The equation is messy so I’ll leave it in this form. Just substitute the x and y values if you need to.

x’(t) = 70 with no air resistance the horizontal velocity doesn’t change with time.
y’(t) = y’(0) – 9.8t = -9.8t
v(y) = r’(t) = √(x’(t)^2 + y’(t)^2) ∠ arctan(y’(t)/x’(t)) m/sec

3. Speed and direction at t= 2 sec.
x(2) = 70*2 = 140 m
y(2) = 100 – 4.9*2^2 = 80.4 m
r(2) = √(x(2)^2 + y(2)^2) ∠ arctan(y(2)/x(2)) m = 161.4 ∠29.9º

x’(2) = 70 m/sec
y’(2) = -9.8*2 = -19.6 m/sec
v(2) = r’(2) = √(x’(2)^2 + y’(2)^2) ∠ arctan(y’(2)/x’(2)) m/sec = 72.7∠−15.6º m/sec

Whew!

Answer 3

r(t)= -1/2g*t^2-Vo*t+ho((for 0 At t=2 s
r(2)=-4.9*4+100 = 80.40 m
V(2) =-9.8*2 =-19.6 m/s downwards and 70m/s horizontally

The velocity vector at t=2 s is(70,-19,6) and its modulus is72.69 m/s