Question

During a test a rocket travels upward at 75 m>s, and when it is 40 m from the ground its engine fails. Determine the maximum height sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m>s2 due to gravity. Neglect the effect of air resistance.

Answer 1

Step-by-step explanation:

Given During a test a rocket travels upward at 75 m>s, and when it is 40 m from the ground its engine fails. Determine the maximum height  sB reached by the rocket and its speed just before it hits the ground. While in motion the rocket is subjected to a constant downward acceleration of 9.81 m>s2 due to gravity. Neglect the effect of air resistance

• Given rocket travels upward with uniform acceleration
• Now ho = 40 m
• Also a = 75 m/s^2
• To = time before engine falls
• Therefore to = √2ho/a
• Therefore velocity of engine will be
• Vo = a x to = √2hoa
• So velocity of rocket will be
• V = vo – gt
• If v = 0 it will be maximum height  
• So t = vo/g
• Or h max = ho + vot – gt^2/2
•               = ho + vo^2 / 2g
•               = ho + hoa/g
•               = ho(1 + a/g)
•               = 40 (1 + 75 / 9.8)
•              = 345.8 m
• The law of conservation of energy  
• So mgh max = mv^2 m / 2 where vm is the speed of rocket before it hits the ground
• So vm = √2gho (1 + a/g)
•           = √2ho(a + g)
•           = √2 x 40 (75 + 9.81)
• So vm = 82.4 m/s

Reference link will be

https://brainly.in/question/82376

Answer 2

The maximum height reached by the rocket is 345.8 m

The speed just before it hits the ground is 82.4 m/s

Step-by-step explanation:

The height reached by the rocket at uniform acceleration is:

h₀ = (at₀²)/2

Where,

h₀ = Height reached before engine fails = 40 m

a = Acceleration of the rocket = 75 m/s²

t₀ = Time traveled before engine fails

Now, the formula becomes,

t₀ = √(2h₀/a)

The velocity before engine fails is given as:

v₀ = a × t₀ = √(2h₀a)

The velocity of the rocket after engine fails is:

v = v₀ - gt

At maximum the velocity, v = 0, then,

0 = v₀ - gt

v₀ = gt

∴ t = v₀/g

Now, the maximum height is given by the formula:

hmax = h₀ + v₀t - gt²/2

hmax = h₀ + v₀²/2g

hmax = h₀ + h₀a/g

hmax = h₀ (1 + a/g)

On substituting the values, we get,

hmax = 40 × (1 + 75/9.81)

∴ hmax = 345.8 m

On applying law of conservation of energy, we get,

mg hmax = mvₐ²/2

vₐ = Speed before rocket hits the ground

vₐ = √(2gh₀(1 + a/g))

vₐ = √(2h₀(a + g))

On substituting the values, we get,

vₐ = √(2 × 40 × (75 + 9.81))

∴ vₐ = 82.4 m/s