Answer:
a) The horizontal velocity of the balloon just before it hits the ground is 6 m/s
b) The magnitude of the vertical velocity of the balloon just before it hits the ground is 98 m/s.
Explanation:
Hi there!
The velocity and position vectors of the water balloon are given by the following equations:
r =(x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v =(v0x, v0y + g · t)
where:
r = position vector at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive) .
v = velocity vector at time t.
a) Please, see the attached figure for a graphic description of the problem.
Considering the origin of the frame of reference as the point of launch, notice that the position vector when the balloon hits the ground is
r1 = (60, -50) m
Then:
r1x = 60 m = v0x · t
r1y = -50 m = 1/2 · (-9.8 m/s²) · t²
(notice that the initial vertical velocity is zero, see figure).
Solving r1y for t:
(-50 m · 2) / -9.8 m/s² = t²
t = 10 s
Now, let´s replace t in the r1x equation and solve it for the horizontal component of the velocity:
60 m = v0x · 10 s
v0x = 60 m / 10 s
v0x = 6 m/s
The initial horizontal component of the velocity is 6 m/s. This velocity is constant because there is no air resistance. Then, just before the balloon hits the ground, it will have a horizontal velocity of 6 m/s.
b) To calculate the vertical component of the velocity when the balloon hits the ground, let´s use the equation of the vertical component of the velocity:
v1y = v0y + g · t
Since v0y = 0
v1y = -9.8 m/s² · (10 s) = -98 m/s
The magnitude of the vertical velocity of the balloon when it hits the ground is 98 m/s.
let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
<u>Answer:</u>
Maximum height reached = 35.15 meter.
<u>Explanation:</u>
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g
So range of projectile,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So
Now we have
So maximum height reached = 35.15 meter.