The velocity, expressed in terms of ME, RE is given as .
The maximum height that the rocket could reach if launched vertically is H = (¹/₂v₀²)/g.
<h3>Maximum height of the rocket</h3>The maximum height reached by the rocket can be modeled using conservation of energy as shown below;
P.Ei + K.Ei = K.Ef + P.Ef
P.E = K.E
mgH = ¹/₂mv²
H = (¹/₂v₀²)/g
Learn more about conservation of energy here: brainly.com/question/166559
Answer:
The speed and direction of the two players immediately after the tackle are 3.3 m/s and 53.4° South of West
Explanation:
given information:
mass of fullback, = 92 kg
speed of full back, = 5.8 to south
mass of lineman, =110 kg
speed of lineman, = 3.6
according to conservation energy,
assume that the collision is perfectly inelastic, thus
initial momentum = final momentum
=
'
m₁v₁ = (m₁+m₂)'
' = m₁v₁/(m₁+m₂)
= (92) (5.8)/(92+110)
= 2.64 m/s
=
'
m₂v₂ = (m₁+m₂)'
' = m₁v₁/(m₁+m₂)
= (110) (3.6)/(92+110)
= 1.96 m/s
thus,
' = √
'²+
'²
= 3.3 m/s
then, the direction of the two players is
θ = 90 - tan⁻¹('/
')
= 90 - tan⁻¹(1.96/2.64)
= 53.4° South of West