1) A stone is dropped from the roof of a building; 2.00 s after that, a second stone is thrown straight down with an initial speed of 25 m/s and the two stones land at the same time.
i) How long did it take the first stone to reach the ground?
ii) How high is the building?
iii) What are the speeds of the two stones just before they hit the ground?
2)A jet fighter is traveling horizontally with a speed of 111 m/s at an altitude of 3.00×102 m, when the pilot accidentally releases an outboard fuel tank.
i) How much time elapses before the tank hits the ground
ii) What is the speed of the tank just before it hits the ground
iii) What is the horizontal distance traveled by the tank.
1a. d1 = 0.5g*t^2 = 4.9*2^2 = 19.6 m. =
Distance fallen by the 1st stone after 2 s.
V1^2 = Vo1^2 + 2g*d
V1^2 = 0 + 19.6*19.6 = 384.16
V1 = 19.6 m/s = Velocity of 1st stone after 2 s.
d1 = d2-19.6 m.
Vo1*t + 0.5g*t^2 = Vo2*t+0.5t^2-19.6
Vo1*t-Vo2*t+4.9t^2-4.9t^2 = -19.6
19.6t-25t = -19.6
-5.4t = -19.6
Tf2 = 3.63 s. = Fall time of 2nd stone.
Tf1 = 3.63 + 2 = 5.63 s. = Fall time of
1st stone.
1b. h = Vo*t + 0.5g*t^2
h = 0 + 4.9(5.63)^2 = 155.3 m.
1c. V1^2 = Vo^2 + 2g*h
V1^2 = 0 + 19.6*155.3 = 3,043.88
V1 = 55.2 m/s.
V2^2 = 25^2 + 19.6*155.3 = 3668.88
V2 = 60.6 m/s.
can u help me solve this also ?
A car from rest and travels for 5.0 s with uniform acceleration of +1.5 m/s2. The driver then applies the brakes, causing a uniform deceleration of -2.0 m/s2. If the brakes are applied for 3.0 s.
i) How fast is the car going at the end of the braking period?
ii) How far has it gone?
iii) Sketch a graph of velocity versus time for the motion.
Hmmm.
1) If y is the stone's height at t, and h is the building height, I get
first stone: y = h - 4.9t^2
2nd stone: y = h - 25(t-2) - 4.9(t-2)^2
i: t = 5.63
the rest is easy...
For the car problem, just use your standard equations of motion.
accelerating: v = at = (1.5)(5) = 7.5
decelerating: v = 7.5 - 2(3) = 1.5
distance: 1/2 (1.5)(25) + 7.5(3) + 1/2 (-2)(9) = 32.25
To solve these problems, we can use the principles of motion and the equations of kinematics.
1) Stone Problem:
i) To find the time it took for the first stone to reach the ground, we can use the equation of motion:
y = (1/2) * g * t^2
The initial velocity of the first stone is 0 m/s since it was dropped. The acceleration due to gravity, g, is approximately 9.8 m/s^2. To find the time, we can rearrange the equation:
t^2 = 2y / g
Plug in the value for y, which is the height of the building.
ii) To find the height of the building, we can use the equation:
y = v0 * t + (1/2) * g * t^2
The initial velocity of the second stone is 25 m/s, and the acceleration due to gravity, g, is approximately 9.8 m/s^2. We need to find the time, which can be rearranged:
y = v0 * t + (1/2) * g * t^2
Let y be the height of the building, t the time it takes to reach the ground, and v0 the initial velocity of the second stone. Rearrange the equation to solve for y.
iii) To find the speeds of the two stones just before they hit the ground, we can use the equation:
vf = v0 + gt
Plug in the values for v0 and t to find the final velocity, vf, for each stone.
2) Jet Fighter Problem:
i) To find the time it takes for the tank to hit the ground, we can use the equation:
y = v0t + (1/2)gt^2
The initial vertical velocity, v0, of the tank is 0 m/s since it was released horizontally, and the acceleration due to gravity, g, is approximately 9.8 m/s^2. Solve for t to find the time it takes.
ii) To find the speed of the tank just before it hits the ground, we can use the equation:
vf = v0 + gt
Plug in the values for v0 and t to find the final velocity, vf, of the tank.
iii) To find the horizontal distance traveled by the tank, we can use the equation:
d = v0t
Plug in the values for v0 and t to find the horizontal distance, d, traveled by the tank.
first stone
h=1/2 g t^2
second stone:
h=1/2 g (t-2)^2
set the equations the same, solve for time t it took the first stone
solve for height of build, h.
for speed, v^2=2gh
2. you are given h, solve for t
h=1/2 g t^2
speed is a combination of horizontal and vertical speed.
vertical=sqrt(2gh)
horizontal given
speed=sqrt(horizontal^2+vertical^2)
distance=horizontal speed*time t