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# Physics Homework Help?

3. A rock is thrown off a vertical cliff at an angle of 51.2o above horizontal. The cliff is 61.3 m high and the ground extends horizontally from the base. The initial speed of the rock is 24.9 m/s. Neglect air resistance. How far, in meters, has the rock moved horizontally at the moment it reaches its maximum height?

How long, in seconds, after being thrown does the rock hit the ground?

How far from the base of the cliff, in meters, is the landing point of the rock?

How high above the ground, in meters, is the rock 2.16 s before it hits the ground?

How far horizontally from the cliff, in meters, is the rock 2.16 s before it hits the ground?

5. The weight of an astronaut plus his space suit on the Moon is only 240 N. How much do they weigh on Earth, in newtons, assuming the acceleration due to gravity on the moon is 1.67 m/s2? What is the mass of the astronaut and his space suit on the Moon, in kilograms? What is the mass of the astronaut and his spacesuit on the Earth, in kilograms?

### 2 Answers

- SpacemanLv 78 months ago
θ = angle at which the rock is thrown = 51.2 ° = 0.894 rad

v = initial velocity of the rock = 24.9 m/s

vx = horizontal velocity of the rock = v·cos θ = 15.6 m/s

vy = vertical velocity of the rock = v·sin θ = 19.4 m/s

g = gravitational acceleration = -9.81 m/s²

H = height of the cliff = 61.3 m

h = maximum height above the cliff = to be determined

(a) How far, in meters, has the rock moved horizontally at the moment

it reaches its maximum height?

When the rock reaches its maximum height above the cliff (h), its velocity is zero.

vy = inititial vertical velocity of the rock = 19.4 m/s

vy(f) = final vertical velocity of the rock = 0 m/s

t = flight time of the rock to its maximum height = to be determined

vy(f) = vy + gt

vy(f) - vy = gt

[vy(f) - vy] / g = t1

t1 = [vy(f) - vy] / g

t1 = [0 m/s - 19.4 m/s] / -9.81 m/s

t1 = 1.98 s time to maximum height

dx(i) = initial horizontal distance = 0 m

ax = horizontal acceleration = 0 m/s²

dx(t) = horizontal distance at time t = to be determined

dx(t) = dx(i) + vx·t + 0.5a·t²

dx(t) = 0 + vx·t + 0

dx(t) = vx·t

dx(t) = 15.6 m/s·1.98 s

dx(t) = 30.9 m horizontal distance when the rock is at maximum height

(b) How long, in seconds, after being thrown does the rock hit the ground?

h = 0 + vy·t + 0.5g·t²

h = 0 m + 19.4 m/s·1.98 s + 0.5(-9.81 m/s²)·(1.98 s)²

h = 19.2 m maximum height above the cliff

y(max) = H + h = 61.3 m + 19.2 m

y(max) = H + h = 80.5 m maximum height above the ground

t2 = √[2(H + h) / |g|]

t2 = √[2(80.5 m) / |-9.81 m/s²|]

t2 = 4.05 s time from maximum height to ground

T = t1 + t2 = total flight time of the rock = to be determined

T = 1.98 s + 4.05 s

T = 6.03 s

(c) How far from the base of the cliff, in meters, is the landing point of the rock?

dx(t) = 15.6 m/s·6.03 s

dx(t) = 94.1 m maximum distance from the base of the cliff

(d) How high above the ground, in meters, is the rock 2.16 s before it hits the ground?

y(t2 -2.16) = distance that the rock is above the ground 2.16 s before it hits = to be determined

t2 - 2.16 = 1.89 s

y(1.89) = distance that the rock is above the ground 2.16 s before it hits = to be determined

1.89 s = √[2·y(1.89)/|g|]

(1.89 s)² = 2·y(1.89)/|g|

|g|(1.89 s)² = 2·y(1.89)

[|g|(1.89 s)²] / 2 = y(1.89)

y(1.89) = [|g|(1.89 s)²] / 2

y(1.89) = [|-9.81 m/s²|(1.89 s)²] / 2

y(1.89) = 17.5 m height above ground 2.16 s before the rock hits

(e) How far horizontally from the cliff, in meters, is the rock 2.16 s before it hits the ground?

dx(T - 2.16) = the rock's horizontal distance 2.16 s before it hits the ground = to be determined

T - 2.16 = 3.87 s

dx(3.87) = 15.6 m/s·3.87 s

dx(3.87) = 60.4 m rock's distance from the cliff at T - 2.16 s

(a) How much do they weigh on Earth, in newtons, assuming the acceleration due to gravity on the moon is 1.67 m/s2?

W(m) = weight of the astronaut and spaesuit on the moon = 240 N

g(m) = lunar gravitational acceleration = 1.67 m/s²

g = Earth gravitational acceleration = 9.81 m/s²

m = mass of the astronaut and spaesuit = to be determined

W(e) = weight of the astronaut and spaesuit on Earth = to be determined

W(m) = m·g(m)

W(m) / g(m) = m

m = W(m) / g(m)

m = 240 N / 1.67 m/s²

m = 144 kg

W(e) = m·g

W(e) = 144 kg·9.81 m/s²

W(e) = 1,410 N

(b) What is the mass of the astronaut and his space suit on the Moon, in kilograms?

(c) What is the mass of the astronaut and his spacesuit on the Earth, in kilograms?

The MASS does NOT change; it remains the SAME whether he is on the moon, Earth,

Mars, Jupiter, or Ceres.

m = 144 kg

- AmyLv 78 months ago
x(t) = x(0) + v(0) * t + a t^2 / 2

Gravity does not affect horizontal speed, and horizontal motion does not affect gravity. Split the initial velocity of the rock into vertical and horizontal components.

Given the initial height, initial vertical velocity, and acceleration of the rock, write a function for height of the rock as a function of time.At what time T does it reach height 0?What is its height at time (T - 2.16s)?Given the initial horizontal speed of the rock, how far does it travel in time T? And how far does it travel in time (T - 2.16s)?

Weight = mass x acceleration due to gravity. Mass is an intrinsic property that does not depend on where you are.

Given the astronaut's weight and g on the moon, what is his mass?

What would he weigh in Earth's gravity?