An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ed by the rocket?
1 answer:
Answer:
18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144
b. The maximum height = 400 feet
c. Attached graph
19) The rocket will reach the maximum height after 4 seconds
20) The rocket hits the ground after 9 seconds
Step-by-step explanation:
* Lets study the rule of motion for an object with constant acceleration
# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time
and a is the acceleration of gravity
# The vertical distances h in x second is h(x) - h(0), where h(0)
is the initial height of the object above the ground
∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial
velocity, a is the acceleration of gravity (32 feet/second²) and x
is the time
18)a.
∵ The value of a = -32 ft/sec² ⇒ negative because the direction
of the motion
is upward
∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16
∴ h(x) = vx - 16x² + h(0)
∴ h(x) = -16x² + vx + h(0) ⇒ proved
* Find the height of the object after x seconds from the ground
∵ h(0) = 144 and v = 128 ft/sec
∴ h(x) = -16x² + 128x + 144
b.
* At the maximum height h'(x) = 0
∵ h'(x) = -32x + 128
∴ -32x + 128 = 0 ⇒ subtract 128 from both sides
∴ -32x = -128 ⇒ ÷ -32
∴ x = 4 seconds
- The time for the maximum height = 4 seconds
- Substitute this value of x in the equation of h(x)
∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet
c. Attached graph
19)
- The object will reach the maximum height after 4 seconds
20)
- When the rocket hits the ground h(x) = 0
∵ h(x) = -16x² + 128x + 144
∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16
∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x
∵ x² - 8x - 9 = 0
∴ (x - 9)( x + 1) = 0
∴ x - 9 = 0 OR x + 1 = 0
∴ x = 9 OR x = -1
- We will rejected -1 because there is no -ve value for the time
* The time for the object to hit the ground is 9 seconds
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