A) Calculate the height (in m) of a cliff if it takes 2.21 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.03 m/s. (No Response) 6.19 m (b) How long (in s) would it take to reach the ground if it is thrown straight down with the same speed? (No Response) 0.571 s

a)Height of the cliff = 6.19 m

b) t = 0.571 s

Explanation:

We need to first calculate the time it takes the rock to go up and come down to the cliff's level first.

Using the equations of motion,

g = -9.8 m/s², u₁ = Initial velocity = 8.03 m/s,

v = velocity at maximum height = 0 m/s

t₁ = time to reach maximum height = ?

a) v₁ = u₁ + gt₁

0 = 8.03 - 9.8t

t₁ = 8.03/9.8 = 0.819 s

Total time of flight for the rock to leave the cliff and reach maximum height = 0.819 s

Height that the rock attains, measured from the top of the cliff to the maximum height, y₁ = ?

y₁ = ut₁ + gt₁²/2 = 8.03(0.819) - 9.8(0.819²)/2 = 3.29m

Total time of flight of the rock, t = t₁ + t₂

Subtracting the time obtained from the first part (0.819 s) from the total time would give the time it took the rock to travel from maximum height reached to the ground.

t₂ = 2.21 - 0.819 = 1.391 s

Let y = total height fallen through = y₁ + (height of the cliff) = y₁ + H = ?

g = 9.8 m/s²

t₂ = 1.391 s

u₂ = initial velocity for 2nd phase of rock motion = velocity at maximum height = 0 m/s

y = u₂t₂ + gt₂²/2

y = 0 + 9.8(1.391²)/2

y = 9.48 m

y = y₁ + H

9.48 = 3.29 + H

H = 9.48 - 3.29 = 6.19 m

b) y = H = 6.19 m, u = 8.03 m/s, t = ?,

y = ut + gt²/2

6.19 = 8.03t + 9.8t²/2

4.9t² + 8.03t - 6.19 = 0

Solving the quadratic eqn

t = -2.21 s or 0.571 s

Since time can only be positive,

t = 0.571 s

(PROVED)

Explanation:

A.

Using equations of motion,

g = -9.81 m/s² (motion against gravity)

u = Initial velocity = 8.03 m/s,

v = velocity at maximum height

= 0 m/s

v = u + g*t

0 = 8.03 - 9.81*t

t₁ = 8.03/9.81

= 0.82 s.

S = u*t + (g*t²)/2

= 8.03*0.82 - (9.81*0.82²)/2

= 3.29 m

Total time of flight, t = t₁ + t₂

t₂ = 2.21 - 0.82

= 1.39 s

Let Y= total height fallen through = S + (height of the cliff)

= S + H

t₂ = 1.39 s

u₂ = initial velocity for motion = velocity at maximum height

= 0 m/s

Y = u₂*t₂ + (g*t₂²)/2

Y = 0 + (9.8*1.39²)/2

= 9.48 m.

Y = S + H

9.48 = 3.29 + H

H = 9.48 - 3.29

= 6.19 m

B.

S1 = H = 6.19 m

u = 8.03 m/s

S1 = u*t + (g*t²)/2

6.19 = 8.03*t + (9.8*t²)/2

4.9*t² + 8.03*t - 6.19 = 0

Using the formular for solving quadratic equations,

t = 0.57 s.

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.