A ball is thrown 107 m upward and then falls back to earth.
Neglecting air resistance, how long will it
be in the air? The acceleration of gravity is 9.8 m/s2 .
Answer in units of s
V^2 = Vo^2 + 2g*h
Vo^2 = -2g*h = -2*-9.8*107 = 2097.2
Vo = 45.8 m/s.
V = Vo + gt = 0 @ max. ht.
Tr = (V-Vo0/g = (0-45.8)/-9.8 = 4.67 s. = Rise time.
Tf = Tr = 4.67 s. = Fall time.
T = Tr + Tf = 4.67 + 4.67 = 9.35 s. =
Time in air.
To determine the time the ball will be in the air, we first need to find the time it takes for the ball to reach its highest point. We can use the kinematic equation:
vf = vi + at
Where:
vf = final velocity (0 m/s at the highest point)
vi = initial velocity (the velocity when the ball is thrown)
a = acceleration (-9.8 m/s^2)
t = time
Since the ball is thrown upward, the initial velocity will be positive. The final velocity at the highest point is 0 m/s.
0 = vi + (-9.8)t
Simplifying the equation:
-9.8t = vi
Now, we can find the time it takes for the ball to reach its highest point.
Next, we can calculate the time it takes for the ball to fall back to the ground. The distance fallen will be equal to the distance thrown upward (107 m) plus the distance fallen from the highest point.
Using the kinematic equation:
d = vi*t + (1/2)at^2
Where:
d = distance (107 m + distance fallen from the highest point)
vi = initial velocity (0 m/s at the highest point)
t = time it takes for the ball to fall back to the ground (which we can assume to be the same time it took to reach the highest point)
a = acceleration due to gravity (-9.8 m/s^2)
Substituting the values into the equation:
107 = 0*t + (1/2)(-9.8)t^2
Simplifying the equation:
4.9t^2 = 107
Dividing by 4.9:
t^2 = 21.83673469
Taking the square root:
t = 4.672 s
Therefore, neglecting air resistance, the ball will be in the air for approximately 4.672 seconds.
To find the time the ball is in the air, you can use the equation of motion:
h = ut + (1/2)at^2
Where:
h = height (107 m)
u = initial velocity (when ball is thrown upwards, its initial velocity is positive)
a = acceleration due to gravity (-9.8 m/s^2, negative because it is acting downwards)
t = time
Since the ball is thrown upward, its initial velocity is positive. So, we can rewrite the equation as:
h = ut - (1/2)gt^2
Substituting the known values:
107 = 0 - (1/2)(-9.8)t^2
Simplifying the equation:
107 = 4.9t^2
Dividing both sides by 4.9:
t^2 = 21.8367347
Taking the square root of both sides to solve for t:
t = √21.8367347
Calculating this value using a calculator or software, we find:
t ≈ 4.67 seconds
Therefore, the ball will be in the air for approximately 4.67 seconds.