Question

A rocket is located on a platform that is 200 feet above a deep canyon. After launching the rocket with an initial velocity of 50 ft/sec, the platform is moved.
a.) What is the max height the rocket will reach?
b.) When will it reach the max height?
c.) When will it be 300 feet off the ground?
d.) How high will it be after 4.2 seconds?
e.) Where will the rocket be after seven seconds?

Answer 1

Answer:

a) Max height it will reach is 327.55 ft above the ground

b) It will reach max height after 5.1 seconds

c) It will be 300 feet off the ground at 12 seconds

d) It will be 305 ft above the ground

e) The rocket will be 247.55 ft above the ground after seven seconds

Step-by-step explanation:

a) Using the equations of kinematics:

v² = v_i² + 2 g Δy

where

• v is the final velocity
• v_i is the initial velocity
• g is the acceleration due to gravity
• Δy is the rocket's displacement

Therefore,

0² = 50² + 2(- 9.8) Δy        

(the negative sign shows that the positive direction is upwards. Gravity acts downwards)

Δy = -(50)² / 2(-9.8)

Δy = 127.55 ft

Thus, the maximum height that the rocket reaches will be

200 ft  + 127.55 ft

= 327.55 ft above the ground

b) Using the equations of kinematics:

Δy = [(v + v_i) / 2 ] × t

t = Δy / [(v + v_i) / 2 ]

t = 127.55 / [(0 + 50) / 2]

t = 5.1 seconds

Therefore, the rocket will reach its maximum height after 5.1 seconds.

c) Using the equations of kinematics:

t₃₀₀ = Δy / [(v + v_i) / 2 ]

t₃₀₀ = 300 / [(0 + 50) / 2]

t₃₀₀ = 12 seconds

Therefore, the rocket will reach 300 feet after 12 seconds

d) Using the equations of kinematics:

Δy = [(v + v_i) / 2 ] × t

Δy = [(0 + 50) / 2] × 4.2

Δy = 105 ft

Therefore, the rocket will be

105 ft + 200 ft

= 305 ft above the ground after 4.2 seconds

e) Using the equations of kinematics:

Δy = [(v + v_i) / 2 ] × t

Δy = [(0 + 50) / 2] × 7

Δy = 175 ft

Therefore,

175 - 127.55 = 47.55 ft

Thus, the rocket will be

47.55 ft + 200 ft

= 247.55 ft above the ground after seven seconds