A person stands on the edge of the roof a building that is 16.5 m tall. The person throws the rock upward with an initial speed of 7.79 m/s. The rock subsequently falls to the ground. Ignore air resistance and use g= 9.8 m/s2 for the acceleration due to gravity.
a. At what speed does the rock hit the ground?
b. How much time does the rock spend in the air?

a) v = 19.60 m/s

b) t = 2.21 s

Explanation:

The time taken to reach the maximum height can be derived from the first equation of motion.

v = u + at 1

given

v = 0 at the maximum height

u = 7.79 m/s

a = -g = -9.8m/s^2

Equation 1 becomes

v = u - gt =0

u = gt

t = u/g

t = 7.79/9.8 = 0.795s

t1 = 0.795s

The maximum height can be derived from;

s = ut + 0.5at^2

Since it's against gravity a = -g and s = h

The equation becomes

h = ut - 0.5gt^2

Substituting the given and derived values, we have

h = 7.79(0.795) - 0.5(9.8 × 0.795^2)

h = 3.096 m

Total maximum height from the ground = 3.096m + 16.5m = 19.596m

H = 19.596m

The final velocity of the rock when it hits the ground can be derived using the fourth equation of motion

v^2 = u^2 + 2as

Motion from the maximum height to the ground.

u = 0 (initial speed at the maximum height is 0)

a = g (acceleration due to gravity) = 9.8m/s^2

s = H = 19.596m

Substituting the values.

v^2 = 0 + 2×9.8× 19.596

v = √( 2×9.8× 19.596)

v = 19.60m/s

Using

v=u+at^2 3

u = 0 and a = g

From equation 3

t^2 = v/g

t =√(v/g)

v = 19.60m/s

t2 = √(19.60/9.8)

t2 = 1.414s

Total time of flight t = t1 + t2 = 0.795s + 1.414s

t = 2.209s

t = 2.21 s

i'm on my way to history not a mistory.