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