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What Initial Velocity would you have to give the Marble to make it hit the ground in 5 s
2 answers:
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Answer:
The distance from the base of the building to the landing site is 154 m.
The total flight time is 3.5 s.
At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.
Explanation:
The equations for the position and velocity vectors of the cat are as follows:
r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = position vector of the cat at time t.
x0 = initial horizontal position.
v0x = initial horizontal velocity.
t = time.
y0 = initial vertical position.
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive).
v = velocity vector of the cat at time t.
Please, see the attached figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the launching point so that x0 and y0 = 0. In a horizontal launch, initially there is no vertical velocity, then, v0y = 0.
When the cat reaches the ground, the position vector of the cat will be r1 (see figure). The vertical component of r1 is -60 m and the horizontal component will be the horizontal distance traveled by the cat (r1x). Then, using the equation of the y-component of the position vector, we can obtain the time of flight and with that time we can obtain the horizontal distance traveled by the cat:
r1y = y0 + v0y · t + 1/2 · g · t²
-60 m = 0 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
- 60 m = -4.9 m/s² · t²
-60 m / - 4.9 m/s² = t²
t = 3.5 s
The cat reaches the ground in 3.5 s
Now, we can calculate the horizontal component of r1:
r1x = x0 + v0 · t
r1x = 0 m + 44 m/s · 3.5 s
r1x = 154 m
The distance from the base of the building to the landing site is 154 m.
The total flight time was already calculated and is 3.5 s.
The velocity vector of the cat when it reaches the ground will be:
v = (v0x, v0y + g · t)
v = (44 m/s, 0 m/s - 9.8 m/s² · 3.5 s)
v = (44 m/s, -34.3 m/s)
The magintude of the vector "v" is calculated as follows:
At the moment of impact, the velocity vector of the cat is v = (44 m/s, -34.3 m/s) and its magnitude is 55.8 m/s.
Answer: a=-2.4525 m/s^2
d=s=190.3 m
Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr
Fr = 25% of weight
W=mg
W=1750*9.81
W=17167.5
Hence
Frictional force is negative as it acts in opposite direction
According to newton second law of motion
F=ma
hence
given
u= 110 km/h
u=110*1000/3600
u=30.55 m/s
to get t we know that final velocity v=0
Answer:
Maximum height of rocket = 2538.74 m
Explanation:
We have equation of motion s = ut + 0.5 at²
For first 5 seconds
s = 0 x 5 + 0.5 x 40 x 5² = 500 m
Now let us find out time after 5 seconds rocket move upward.
We have the equation of motion v = u + at
After 5 seconds velocity of rocket
v = 0 + 40 x 5 = 200 m/s
After 5 seconds the velocity reduces 9.8m/s per second due to gravity.
Time of flying after 5 seconds
Distance traveled in this 20.38 s
s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m
Maximum height of rocket = 500 +2038.74 = 2538.74 m