A bullet is fired vertically upward from a gun and reaches a height of 7000 ft. Find its initial velocity. How long does it take to reach its maximum height? How long is it in flight?
at the top, vertical velocity is zero
h=vi*t-1/2 g t^2
7000=vi*t-16t^2
this is a quadratic, solve for time.
Initial KE=final PE
1/2 m v^2=mgh solve for v, the initial velocity
how long is the flight? Twice the time to the top.
To find the initial velocity of the bullet, we'll use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0, since the bullet reaches its maximum height and momentarily stops)
u = initial velocity (what we want to find)
a = acceleration due to gravity (-32 ft/s^2, assuming downward acceleration)
s = displacement (7000 ft, maximum height reached by the bullet)
Using the equation, we can rearrange it to solve for u:
0 = u^2 + 2(-32)(7000)
0 = u^2 - 448000
Taking the square root of both sides, we get:
u = ±√448000
Since the bullet is fired upward, the initial velocity has to be positive:
u = √448000
u ≈ 669.55 ft/s
Therefore, the initial velocity of the bullet is approximately 669.55 ft/s.
To find how long it takes to reach its maximum height, we can use the equation:
v = u + at
Since the final velocity is 0 (at maximum height), we have:
0 = u + (-32)t
Rearranging the equation, we get:
t = u/32
t ≈ 669.55/32
t ≈ 20.92 seconds
Therefore, it takes approximately 20.92 seconds for the bullet to reach its maximum height.
To find how long the bullet is in flight, we can use the equation:
s = ut + (1/2)at^2
Since the bullet reaches a height of 7000 ft and comes back down to the ground, the displacement, s, is 0. We can rearrange the equation to solve for t:
0 = ut + (1/2)(-32)t^2
0 = ut - 16t^2
Factoring out t, we have:
t(u - 16t) = 0
This equation has two solutions: t = 0 (when the bullet is initially fired) and u - 16t = 0 (when the bullet reaches the ground).
Solving u - 16t = 0 for t, we get:
u = 16t
t = u/16
t ≈ 669.55/16
t ≈ 41.85 seconds
Therefore, the bullet is in flight for approximately 41.85 seconds.
To find the initial velocity of the bullet, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Here,
vf = final velocity (which is 0 when the bullet reaches its maximum height)
vi = initial velocity (what we want to find)
a = acceleration due to gravity (approximately -32.2 ft/s^2)
d = displacement (7000 ft)
Substituting the given values into the equation:
0 = vi^2 + 2(-32.2)(7000)
Simplifying the equation, we have:
0 = vi^2 - 451400
To solve for vi, we can take the square root of both sides:
vi = sqrt(451400)
Calculating the square root, we find:
vi ≈ 671.43 ft/s
Therefore, the initial velocity of the bullet is approximately 671.43 ft/s.
To find the time it takes for the bullet to reach its maximum height, we can use the equation:
vf = vi + at
Here,
vf = final velocity (0 ft/s)
vi = initial velocity (671.43 ft/s, which is already found)
a = acceleration due to gravity (-32.2 ft/s^2)
t = time (what we want to find)
Substituting the given values into the equation:
0 = 671.43 + (-32.2)t
Simplifying the equation, we have:
671.43 = 32.2t
Dividing both sides of the equation by 32.2, we find:
t ≈ 20.84 s
Therefore, it takes approximately 20.84 seconds for the bullet to reach its maximum height.
To find the total time the bullet is in flight, we can double the time it takes to reach the maximum height:
Total time = 2 * time to reach maximum height
Total time = 2 * 20.84 s
Total time ≈ 41.68 s
Therefore, the bullet is in flight for approximately 41.68 seconds.