Answer:
a) x = v₀ₓ t
, y = t - ½ g t²
Explanation:
This is a projectile launch problem, let's use Newton's second law on each axis
X axis
F = m a
Since there is no acceleration on the x axis, the force on this axis is zero
Y Axis
-W = m
-m g = m
= -g
In this axis the acceleration is the acceleration of gravity
Now we can use science to find the position of the body on each axis
X axis
x = v₀ₓ t + ½ a t²
As the acceleration on this axis is zero
x = v₀ₓ t
Y Axis
y = t + ½
t²
The acceleration on this axis is –g
y = t - ½ g t²
B) to find the maximum value of distance r
r =√ x² + y²
r = √( v₀ₓ² t² + ( t + ½ g t²)²
We can find the maximum value of r using time respect derivatives
dr / dt = 0
0 = ½ 1/√( v₀ₓ² t² + ( t + ½ g t²)² (v₀ₓ² 2t + 2 (
t - ½ g t²)(
- ½ g2t)
We simplify this expression
0 = v₀ₓ² 2t + 2 ( t - ½ g t²) (
- ½ g2t)
-v₀ₓ² t =( t - ½ g t²) (
- ½ g2t)
-v₀ₓ² t = ² t -3/2 gt²
+ 1/2 g² t³
½ g² t²- 3/2 g t =
² + v₀ₓ²
Let's use trigonometry to find go and vox
sin θ = / v₀
cos θ = vox / v₀
= v₀ sin θ
v₀ₓ = v₀ cos θ
We replace
½ g² t² -3/2 g t = v₀ (sin² θ + cos² θ)
g t² - 3v₀ sin θ t = 2 v₀/g
The time is maximum for the angle is zero