An explosion causes debris to rise vertically with an initial velocity of 75 ft/sec.

a. In how many seconds does it attain maximum height?
b. What is the maximum height?
c. What is the speed of the debris as it reaches 12 ft on the way back to the ground?
d. What is the velocity of the debris at 20 ft on the way up

Easy way is to assume that total energy is constant (of course debris might have a lot of air drag but there is not much we can do about that.

mass m (will cancel everywhere in this problem)

(1/2) m v^2 + m g h = total energy
at start h = 0 and v = 75
so
(1/2)(75)^2 = total energy/m = 2813
in obsolete English units as here g = 32 ft/s^2
so at 12 ft
(1/2) v^2 = 32(12)
solve for v
and at 20 ft
(1/2) v^2 = 32 (20) solve for v

note that the speed is the same upwards and downwards at the same height, just the direction changes.

at top

(1/2) v^2 = 0
so
32 h = 2813

at 12 feet
(1/2)v^2 + 32 (12) = 2813

at 20 feet
(1/2) v^2 + 32 (20) = 2813

a)

Height(t) = -16t^2 + 75t
Height ' (t) = -32t + 75 = 0 for a max of Height
-32t = -75
t = 75/32 seconds or appr 2.34 seconds

b) Height(75/32) = -16(75/32)^2 + 75(75/32) = 87.89 ft

c) when Height = 12
12 = -16t^2+ 75t
16t^2 - 75t + 12 = 0
t = 4.5216 sec (on the way down) or
t = .1659 seconds (on the way up)

velocity(t) = height ' (t)
= -32t + 75 , so we want it on the way down
velocity(4.5216) = -32(4.5216)+75 = -69.69 ft/sec

d)
same method as in c)
16t^2 - 75t + 20 = 0
t = 4.40 seconds (on the way down), or
t = .284 secpmds (on the way up

velocity(.284) = -32(.284)+75 = 65.92 ft/sec

To answer these questions, we can use the equations of motion for vertical motion under the influence of gravity. These equations are:

1. h(t) = h₀ + v₀t - (1/2)gt²
2. v(t) = v₀ - gt

Where:
- h(t) is the height at time t
- h₀ is the initial height
- v₀ is the initial velocity
- g is the acceleration due to gravity (approximately 32 ft/sec²)
- t is the time

Let's solve each part of the problem step by step:

a. In how many seconds does the debris attain maximum height?

To find the time at which the debris attains maximum height, we need to determine when its vertical velocity becomes zero. Since the debris rises vertically, its initial velocity of 75 ft/sec is positive. We can set the equation for vertical velocity, v(t), equal to zero and solve for t:

v₀ - gt = 0
75 - 32t = 0
32t = 75
t = 75 / 32
t ≈ 2.34 seconds

Therefore, the debris takes approximately 2.34 seconds to attain maximum height.

b. What is the maximum height?

To find the maximum height, we can substitute the time obtained in part (a) into the equation for height, h(t):

h(t) = h₀ + v₀t - (1/2)gt²
h(t) = 0 + 75(2.34) - (1/2)(32)(2.34)²
h(t) ≈ 175.2 ft

Therefore, the maximum height attained by the debris is approximately 175.2 ft.

c. What is the speed of the debris as it reaches 12 ft on the way back to the ground?

To find the speed of the debris at a particular height, we first need to determine the time it takes to reach that height. We can set the equation for height, h(t), equal to the desired height and solve for t:

h(t) = h₀ + v₀t - (1/2)gt²
12 = 0 + 75t - (1/2)(32)t²
16t² - 150t + 24 = 0

Solving this quadratic equation will give us the time it takes to reach 12 ft. Once we find the time, we can substitute it into the equation for velocity, v(t), to get the speed.

d. What is the velocity of the debris at 20 ft on the way up?

Similar to part (c), we need to find the time it takes for the debris to reach the height of 20 ft. We can set the equation for height, h(t), equal to 20 and solve for t:

h(t) = h₀ + v₀t - (1/2)gt²
20 = 0 + 75t - (1/2)(32)t²
16t² - 150t + 80 = 0

Solving this quadratic equation will give us the time it takes to reach 20 ft. Once we find the time, we can substitute it into the equation for velocity, v(t), to get the velocity.