E. An object is thrown from the top of a tower with
1 answer:
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Answer:
A) The maximum height reached by the projectile is 34.3 m.
B) The total time in the air is 5.29 s.
C) The range of the projectile is 144 m.
D) The speed of the projectile 1.80 s after firing is 28.4 m/s.
Explanation:
Please, see the attached figure for a better understanding of the problem.
The position and velocity vectors of the projectile at time "t" are as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time "t"
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
v = vector position at time t
Let´s place the origin of the frame of reference at the launching point so that x0 and y0 = 0.
A) At the maximum height, the vertical component of the velocity is 0 (see figure). Then, using the equation for the y-component of the velocity vector, we can obtain the time at which the projectile is at its maximum height:
vy = v0 · sin α + g · t
0 = 37.6 m/s · sin 43.6° - 9.8 m/s² · t
- 37.6 m/s · sin 43.6° / -9.8 m/s² = t
t = 2.65 s
The height of the projectile at this time will be the maximum height. Then, using the equation of the y-component of the vector position:
y = y0 + v0 · t · sin α + 1/2 · g · t² (y0 = 0)
y = 37.6 m/s · 2.65 s · sin 43.6° - 1/2 · 9.8 m/s² · (2.65)²
y = 34.3 m
The maximum height reached by the projectile is 34.3 m.
B) When the projectile reaches the ground, the y-component of the position vector is 0 (see vector "r final" in the figure). Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
0 = 37.6 m/s · t · sin 43.6° - 1/2 · 9.8 m/s² · t²
0 = t · (37.6 m/s · sin 43.6° - 1/2 · 9.8 m/s² · t) (t = 0, the initial point)
0 = 37.6 m/s · sin 43.6° - 1/2 · 9.8 m/s² · t
- 37.6 m/s · sin 43.6° /- 1/2 · 9.8 m/s² = t
t = 5.29 s
The total time in the air is 5.29 s.
C) Having the total time in the air, we can calculate the x-component of the vector "r final" (see figure) to obtain the horizontal distance traveled by the projectile:
x = x0 + v0 · t · cos α
x = 0 m + 37.6 m/s · 5.29 s · cos 43.6°
x = 144 m
The range of the projectile is 144 m.
D) Let´s find the velocity vector at that time:
v = (v0 · cos α, v0 · sin α + g · t)
vx = v0 · cos α
vx = 37.6 m/s · cos 43.6°
vx = 27.2 m/s
vy = v0 · sin α + g · t
vy = 37.6 m/s · sin 43.6° - 9.8 m/s² · 1.80 s
vy = 8.29 m/s
Then, the vector velocity at t = 1.80 s will be:
v = (27.2 m/s, 8.29 m/s)
The speed is the magnitude of the velocity vector:
The speed of the projectile 1.80 s after firing is 28.4 m/s.
Answer:
the cart's speed at point B is 15.72 ft/s
Explanation:
Given the data in the question;
The car travels from point A to C in 3.00 s, its average acceleration will be;
= [
-
] / Δt
is 17.4 ft/s,
is 13.2 ft/s and Δt is 3.00 s
so we substitute
= [17.4 - 13.2] / 3
= 4.2 / 3
= 1.4 ft/s²
so average acceleration of the cart between the points A and B is 1.4 ft/s²
The instantaneous value of the velocity of the cart at point B will be;
= Δv / Δt
now substitute [ -
] for Δv and t' for Δt
= [
-
] / t'
=
-
( t' )
so we substitute 17.4 ft/s for , 1.20 s for t' and
= 1.4 ft/s²
= 17.4 - (1.4 × 1.20)
= 17.4 - 1.68
= 15.72 ft/s
Therefore, the cart's speed at point B is 15.72 ft/s
Explanation:
Formula for final volume of chamber if the partition is ruptured will be as follows.
= 1.5 + 1.5
= 3.0
As mass remains constant then the specific volume at this state will be as follows.
=
= 1.5
Now, at final temperature = 300 F according to saturated water tables.
Hence, we obtained and the state is in wet condition.
1.5 =
= 0.229
Now, the final pressure will be the saturation pressure at = 300 F
and, =
= 66.985 psia
Formula to calculate internal energy at the final state is as follows.
=
= 920.56 Btu
Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.