A Projectile IS FIRED FROM THE TOP OF A BUILDING WITH AN INITIAL VELOCITY STRAIGH UPWARD AT 40.0 M/SEC . THE TOP OF THE BUILDING IS 50.0 METERS ABOVE THE GROUND.

A) THE MAXIMUN HEIGHT OF THE PROJECTILE ABOVE THE BUILDING.
B) THE TIME FOR THE PROJECTILE TO REACH THE TOP.
C) THE VELOCITY OF THE PROJECTILE AS HIT THE GROUND
D) THE TOTAL TIME THE PROJECTILE IS IN THE AIR

A. Vf^2 = Vo^2 + 2gh.

h = (Vf^2 _ Vo^2) / 2g,
h = (0 - (40)^2) / -19.6 = 81.63m.

B. t(up) = (Vf - Vo) / g,
t(up) = (0 - 40) / -9.8 = 4.08s.

C. h = 50 + 81.63 = 131.63m. above ground.
Vf^2 = Vo^2 + 2g*h.
Vf^2 = 0 + 19.6*131.6 = 2580,
Vf = 50.8m/s.

D. t(dn) = (Vf - Vo) / g,
t(dn) = (50.8 - 0) / 9.8 = 5.18s.
T = t(up) + t(dn) = 4.05 + 5.18 = 9.23s
in air.

Maximum hieght is=46.9m

(B)answer=3.064m
time of flit is =6.128m

To find the answers to these questions, we can use the equations of motion for a projectile. Let's break it down step by step:

A) The maximum height of the projectile can be found using the equation:

hmax = (v₀²) / (2g)

where v₀ is the initial velocity (40.0 m/s) and g is the acceleration due to gravity (9.8 m/s²).

Plug in the values:

hmax = (40.0 m/s)² / (2 * 9.8 m/s²)

hmax = 1600 m²/s² / 19.6 m/s²

hmax = 81.63 m

Therefore, the maximum height of the projectile above the ground is 81.63 meters.

B) To find the time it takes for the projectile to reach the top, we can use the equation:

h = v₀t - (1/2)gt²

Since the projectile reaches its maximum height at the top, the final height at the top is 50.0 meters and the initial height is 0 meters.

50.0 m = (40.0 m/s)t - (1/2)(9.8 m/s²)t²

Simplifying the equation, we get:

4.9t² - 40t + 50 = 0

Using the quadratic formula, we find two possible solutions for t:

t = (40 ± √(40² - 4 * 4.9 * 50)) / (2 * 4.9)

After calculating, we find t = 8.163 and t = 1.837.

Therefore, the time it takes for the projectile to reach the top is 8.163 seconds.

C) The velocity of the projectile as it hits the ground can be found using the equation:

v = v₀ - gt

where v is the final velocity of the projectile, v₀ is the initial velocity (40.0 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is the total time in the air.

We already know that the total time in the air is twice the time it takes to reach the top, so t = 2 * 8.163 = 16.326 seconds.

v = 40.0 m/s - (9.8 m/s²)(16.326 s)

v = 40.0 m/s - 160.1976 m/s

v ≈ -120.2 m/s

The negative sign indicates that the velocity is in the opposite direction to the initial velocity. Therefore, the velocity of the projectile as it hits the ground is approximately -120.2 m/s.

D) Finally, the total time the projectile is in the air can be obtained by doubling the time to reach the top:

Total time = 2 * 8.163 s = 16.326 seconds

Therefore, the total time the projectile is in the air is 16.326 seconds.