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

Question

3 years ago

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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

Physics

2 answers:

Setler79 [48] · Answer 1 · 2022-02-08T21:01:25+03:00

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.

DedPeter [7] · Answer 2 · 2022-02-08T18:36:30+03:00

Answer:

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 = ?,