It takes 3 s for a rock to hit the ground when it is thrown straight up from a cliff with an initial velocity of 8.63 m/s. how long a time would it take to reach the ground if it is thrown straight down with the same speed?

Explanation:

h = height of the cliff

Consider upward direction as positive and downward direction as negative

Consider the motion of rock thrown straight up :

Y = vertical displacement = - h

v₀ = initial velocity = 8.63 m/s

a = acceleration = - 9.8 m/s²

t = time taken to hit the ground = 3 s

Using the equation

Y = v₀ t + (0.5) a t²

- h = (8.63) (3) + (0.5) (- 9.8) (3)²

h = 18.21 m

Consider the motion of rock thrown down :

Y' = vertical displacement = - 18.21

v'₀ = initial velocity = - 8.63 m/s

a' = acceleration = - 9.8 m/s²

t' = time taken to hit the ground = ?

Using the equation

Y' = v'₀ t' + (0.5) a' t'²

- 18.21 = (- 8.63) t' + (0.5) (- 9.8) t'²

t' = 1.2 s

more than, less

less than, more

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.