A ball is thrown vertically upward with velocities of 18 m/s. Two seconds later another ball is thrown upwards with a velocity of 13.5 m/s. At what position above the ground will they meet?
For the 1st ball, the time of the upward motion
v=v₀-gt
v=0
0=v₀-gt
t= v₀/g =18/9.8=1.83 s.
The height of the 1st ball
h= v₀t-gt²/2=18•1.83 - 9.8•(1.83)²/2= =16.53 m
The time of the 1st ball downward motion before the 2nd ball begins to move is
Δt=2 s -1.83 s = 0.17 s
The 1st ball during 0.17 s covered the distance
Δh=g(Δt)²/2 =9.8•0.17²/2 = 0.14 m.
Its downward velocity is
v₀₁ = gΔt=9.8•0.17 =1.67 m/s
Now, two balls begin to move:
the 1st ball moves downward with initial velocity v₀₁ =1.67 m/s,
the 2nd ball moves upward with initial velocity v₀₂=13.5 m/s.
The distance separated them is
h₀ = h- Δh = 16.53 – 0.14 = 16.39 m.
Before the meeting, the 1st ball covered h₁=v₀₁t+gt²/2,
and the 2nd ball covered the distance h₂=v₀₂t-gt²/2.
h₀ = h₁+h₂=v₀₁t+gt²/2 + v₀₂t-gt²/2=
=(v₀₁+ v₀₂)t
t= h₀/(v₀₁+ v₀₂)=16.39/(13.5+1.67) =
=1.08 s.
The position above the ground is
h₂=v₀₂t-gt²/2 =13.5•1.08 – 9.8•1.08²/2 =
= 14.58 -5.72 = 8.86 m
A 1500 kg automobile is traveling up to 20 degree incline at a speed of 6 m/s. If the driver wishes to stop his car in a distance of 5m, determine the frictional force at pavement which must be supplied by rear wheels.
you can just solve to see when the heights are equal:
18t-4.9t^2 = 13.5(t-2)-4.9(t-2)^2
t = 1.08
proceed from there as above
To find the position where the two balls meet, we can start by calculating the time it takes for each ball to reach its highest point.
Let's call the time when the first ball is thrown "t1" and the time when the second ball is thrown "t2". From the given information, t1 = 0 (as the first ball is thrown instantly) and t2 = 2 seconds.
For both balls, the highest point will be reached when their vertical velocities become zero, assuming no air resistance.
Let's calculate the time it takes for the first ball to reach its highest point:
Using the equation: vf = vi + at
Where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken.
For the first ball:
vf = 0 (at the highest point, velocity becomes zero)
vi = 18 m/s (as given)
a = -9.8 m/s² (acceleration due to gravity, acting downward)
0 = 18 - 9.8t₁
Rearranging the equation:
9.8t₁ = 18
t₁ = 18 / 9.8
t₁ ≈ 1.84 seconds
Now, let's calculate the time it takes for the second ball to reach its highest point:
For the second ball:
vf = 0
vi = 13.5 m/s
a = -9.8 m/s²
0 = 13.5 - 9.8t₂
Rearranging the equation:
9.8t₂ = 13.5
t₂ = 13.5 / 9.8
t₂ ≈ 1.38 seconds
The second ball takes approximately 1.38 seconds to reach its highest point.
Since the first ball is thrown 2 seconds before the second ball, we need to calculate the time difference between them:
Δt = t₂ - t₁
Δt = 1.38 - 1.84
Δt ≈ -0.46 seconds
The negative value indicates that the first ball reaches its highest point 0.46 seconds before the second ball. Now we can find the position where they meet.
Using the equation: h = vt + (1/2)at²
Where h is the height, v is the initial velocity, t is the time, and a is the acceleration.
For the first ball:
h₁ = 18t - 4.9t₁²
Substituting the values:
h₁ = 18(0.46) - 4.9(0.46)²
h₁ ≈ 4.126 meters
Therefore, the two balls will meet approximately 4.126 meters above the ground.