1) A certain aircraft has a liftoff speed of 111 km/h.
(a) What minimum constant acceleration does the aircraft require if it is to be airborne after a takeoff run of 206 m?
(b) How long does it take the aircraft to become airborne?
2) A ball is thrown vertically upward with a speed of 36.0 m/s.
(a) How high does it rise?
(b) How long does it take to reach its highest point?
(c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is its velocity when it returns to the level from which it started?
a. Vf^2= 2*acceleration*distance....solve for a
b. vf=a*t solve for time t.
2.
a. final PE= initial KE
mgh=1/2 mv^2 or h=1/2 v^2/g
1) (a) To find the minimum constant acceleration, we can use the equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity (0 km/h), vi is the initial velocity (111 km/h), a is the acceleration, and d is the distance (206 m).
Converting the speeds to m/s, we have:
vf = 0 km/h = 0 m/s
vi = 111 km/h = 111 m/s
d = 206 m
Plugging these values into the equation, we have:
0^2 = (111^2) + 2a(206)
Simplifying the equation:
0 = 12321 + 412a
Rearranging the equation and solving for a:
412a = -12321
a = -12321 / 412
a ≈ -29.94 m/s^2
So, the minimum constant acceleration required for the aircraft to be airborne is approximately -29.94 m/s^2 (negative sign indicates deceleration).
(b) To find the time it takes for the aircraft to become airborne, we can use the equation:
vf = vi + at
Where vf is the final velocity (0 m/s), vi is the initial velocity (111 m/s), a is the acceleration, and t is the time.
Plugging in the known values:
0 = 111 + (-29.94)t
Simplifying the equation:
29.94t = 111
t = 111 / 29.94
t ≈ 3.71 s
So, the aircraft takes approximately 3.71 seconds to become airborne.
2) (a) To find how high the ball rises, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity (0 m/s), vi is the initial velocity (36.0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and d is the displacement (which we need to determine).
Plugging in the known values:
0^2 = (36.0^2) + 2(-9.8)d
Simplifying the equation:
0 = 1296 - 19.6d
Rearranging the equation and solving for d:
19.6d = 1296
d = 1296 / 19.6
d ≈ 66.12 m
So, the ball rises to a height of approximately 66.12 meters.
(b) To find the time it takes for the ball to reach its highest point, we can use the equation:
vf = vi + at
Where vf is the final velocity (0 m/s), vi is the initial velocity (36.0 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time we need to determine.
Plugging in the known values:
0 = 36.0 + (-9.8)t
Simplifying the equation:
9.8t = 36.0
t = 36.0 / 9.8
t ≈ 3.67 s
So, it takes approximately 3.67 seconds for the ball to reach its highest point.
(c) When the ball reaches its highest point, its vertical velocity becomes 0. From this point, the ball starts falling back down to the ground, so the total time it takes for the ball to hit the ground is twice the time it takes to reach the highest point:
Total time = 2 * time to reach highest point
Total time = 2 * 3.67 s
Total time ≈ 7.34 s
So, it takes approximately 7.34 seconds for the ball to hit the ground after reaching its highest point.
(d) When the ball returns to the level from which it started, its velocity will have the same magnitude as its initial velocity but in the opposite direction. Therefore, the velocity will be -36.0 m/s.
So, the velocity of the ball when it returns to the level from which it started is -36.0 m/s.
To answer these questions, we need to use the equations of motion for uniformly accelerated motion.
For the first question:
(a) To find the minimum constant acceleration, we can use the equation for distance traveled during uniformly accelerated motion:
d = (v^2 - u^2) / (2a)
where d is the distance traveled, v is the final velocity, u is the initial velocity, and a is the constant acceleration.
Rearranging the equation, we get:
a = (v^2 - u^2) / (2d)
Plugging in the given values:
v = 111 km/h = 111 * 1000 / 3600 m/s (conversion from km/h to m/s)
u = 0 m/s (since the initial velocity is zero when the aircraft is at rest)
d = 206 m
Now, substitute the values into the equation to find the minimum acceleration.
(b) To find the time it takes for the aircraft to become airborne, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time taken.
Plugging in the given values:
v = 111 km/h = 111 * 1000 / 3600 m/s
u = 0 m/s
a is the minimum acceleration obtained from part (a)
Rearranging the equation, we get:
t = (v - u) / a
Substitute the values into the equation to find the time.
For the second question:
(a) To find how high the ball rises, we can use the equation for the maximum height reached in vertical motion:
h = (v^2 - u^2) / (2g)
where h is the maximum height reached, v is the final velocity, u is the initial velocity, and g is the acceleration due to gravity.
Plugging in the given values:
v = 36.0 m/s
u = -36.0 m/s (taking upward direction as positive)
g = 9.8 m/s^2
Substitute the values into the equation to find the maximum height.
(b) To find the time taken to reach the highest point, we can use the equation:
v = u + gt
where t is the time taken.
Plugging in the given values:
v = 0 m/s (at the highest point, the velocity becomes zero)
u = 36.0 m/s
g = -9.8 m/s^2 (since the acceleration due to gravity acts in the downward direction)
Rearranging the equation, we get:
t = (v - u) / g
Substitute the values into the equation to find the time.
(c) To find the time taken for the ball to hit the ground after reaching the highest point, we can use the equation:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Plugging in the given values:
h = maximum height obtained in part (a)
u = 0 m/s
g = 9.8 m/s^2
Rearranging the equation, we get a quadratic equation with one unknown t, and solve it to find the time.
(d) To find the velocity when the ball returns to the level from which it started, we can use the equation:
v = u + gt
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Plugging in the given values:
u = -36.0 m/s
g = 9.8 m/s^2
t is the time obtained in part (c)
Substitute the values into the equation to find the velocity.
By following these steps, you should be able to find the answers to the given questions.