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# physics help please!?

A cliff diver stands 45.0 m from the edge of a cliff on a stretch of land that angles downward as it approaches the edge. The angle of the top of the cliff is θ = 19.0°. The cliff diver runs towards the edge of the cliff with a constant speed, and reaches the edge of the cliff in a time of 6.40 s. After running straight off the edge of the cliff (without jumping up), the diver falls h = 24.0 m before hitting the water.

(a) After leaving the edge of the cliff, how much time elapses before the diver enters the water?

(b) How far does the diver travel horizontally from the face of the cliff before hitting the water?

### 3 Answers

- 胡雪8°Lv 72 years agoFavorite Answer
(a)

Take g = 9.81 m/s

Speed that the diver leaving the cliff, u

= Speed that the diver runs towards the edge

= (45.0 m) / (6.40 s)

= 7.03 m/s

Consider the vertical motion (uniform acceleration motion, downward quantities to be positive) :

Initial speed, u = 7.03 sin19.0° m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 24.0 m

s = ut + (1/2)at²

24.0 = (7.03 sin19.0°)t + (1/2)(9.81)t²

(1/2)(9.81)t² + (7.03 sin19.0°)t - 24.0 = 0

t = {-(7.03 sin19.0°) ± √[(7.03 sin19.0°)² + 4(1/2)(9.81)(24/0)]} / [2(1/2)(9.81)]

t = 1.99 s or t = -2.46 s (rejected)

Time taken = 1.99 s

(b)

Horizontal distance

= (7.03 cos19.0° m/s) × (1.99 s)

= 13.2 m

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- 2 years ago
The plan:

i) Find his velocity down the slope as he runs towards the edge.

ii) Resolve that initial velocity into its vertical and horizontal components

iii) Kinematics on the vertical motion to find the time taken to hit the water

iv) Use the time (t) fron (iii) with the constant horizontal component from (ii) to find horizontal distance.

i) Velocity down slope = displacement/time = (45.0 m / 6.40 s) = 7.03 m/s

ii) Vertical component of launch speed = 7.03sin(19.0°) = 2.29 m/s

Horizontal component of launch speed = 7.03cos(19.0°) = 6.65 m/s

iii) Kinematics on the vertical component of motion. The symbols used vary from place to place. I use the “suvat” symbols: s = displacement; u = initial velocity; v = final velocity; a = acceleration; t = time

I choose arbitrarily to say that "downwards is the positive direction". What do we know here?

s = 24.0 m

u = 2.29 m/s

v = {not interested}

a = 9.81 m/s² {Acceleration due to gravity downwards}

t = ? {That is what we want to know}

Equation choice: I could use s = ut + ½at² to find (t) directly, but I am too full of Sunday lunch to fool about with quadratics, so I'll find (v) first...

v² = u² + 2as

v² = 2.29² + 2 * 9.81 * 24.0

v² = 476.12

v = 21.8 m/s

Now: v = u + at

21.8 = 2.29 + 9.81t

(a) t = 1.99 seconds

iv) So travelling with a horizontal component of 6.65 m/s, how far does he get in the 1.99 s it takes to hit the water?

(b) distance = speed * time = (6.65 m/s * 1.99 s) = 13.2 m

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- Randy PLv 72 years ago
"cliff diver stands 45.0 m from the edge of a cliff...reaches the edge of the cliff in a time of 6.40 s."

That tells you his initial velocity v = distance/time.

And you're told it's at an angle θ relative to the horizontal. So you know his horizontal velocity vx = v cos(θ) and his initial vertical velocity vy = v sin(θ).

a) Accelerated motion, d = vy * t + (1/2)at^2. You know d, vy, and a. Solve for t. Note that d and a are downward and vy is upward, so you should use positive signs for one and negative for the other. Doesn't matter whether you use positive for up or for down, just be consistent.

b) Constant-velocity motion, d = vx * t.

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