Word Problem with Muzzle Velocity

Question

From an airplane whose altitude is 2064 feet a bullet is fired with an upward muzzle velocity of 3200 feet per second. When will the bullet's altitude be 8400 feet? Explain the two solutions

Answer 1

x=x(0)+v(0)t+.5at^2

8400=2064+3200t-16t^2

t=2 and 198 seconds, on the way up and down

Answer 2

h(t) = altitude as function of time

 

Assume up is positive, down is negative.

 

h(t) = h₀ + v₀t + (1/2)at²

 

a = -(acceleration of gravity) = -g = -32 ft/s² (acceleration is downward even though the object was shot upward)

 

h₀ = initial height = 2064 ft

 

v₀ = initial upward velocity = 3200 ft/s

 

h(t) = 2064 + 3200t - 16t²

 

When h = 8400,

 

8400 = 2064 + 3200t - 16t²

 

-16t² + 3200t - 6336 = 0

 

Divide through by -16 to simplify a little and get

 

t² - 200t + 396 = 0

 

t = {200 ± √[(-200)² - 4(1)(396)]} / 2 = (200 ± 196) / 2 = 2 or 198

 

Since there are two valid solutions in time, the lower time is the time it reaches 8400 ft on the way up, and the higher time is the time it reaches 8400 ft on the way down.

 

It takes 2 seconds from launch to reach 8400 ft going up, and it takes 198 seconds from launch to reach 8400 ft coming back down.