A juggler tosses a ball h = 2 m into the air with one hand and catches it with the other hand at the same level. Neglecting air resistance, how long is the ball in the air?
0.5g*t^2 = 2
4.9t^2 = 2
t^2 = 0.408
Tf = 0.639 s. = Fall time.
Tr = Tf = 0.639 s. = Rise time.
Tr+Tf = 0.639 + 0.639 = 1.28 s. = Time
in air.
To solve this problem, we can use the kinematic equation for vertical motion:
y = h0 + V0*t - (1/2)*g*t^2
Where:
y = final height (0 m)
h0 = initial height of the ball (2 m)
V0 = initial vertical velocity (unknown)
g = acceleration due to gravity (-9.8 m/s^2)
t = time in seconds (unknown)
Since the ball is thrown upward, the initial vertical velocity (V0) is positive.
To find the time the ball is in the air, we need to solve for t when y = 0.
0 = h0 + V0*t - (1/2)*g*t^2
Rearranging the equation:
(1/2)*g*t^2 - V0*t - h0 = 0
We can now solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = (1/2)*g
b = -V0
c = -h0
Plugging in the values:
a = (1/2)*(-9.8) = -4.9
b = -V0
c = -2
t = (-(-V0) ± √((-V0)^2 - 4*(-4.9)*(-2))) / (2*(-4.9))
Simplifying the equation:
t = (V0 ± √(V0^2 + 39.2)) / 9.8
Since the ball is caught at the same level, the final height is the same as the initial height (h0 = 2 m), which means y = 0.
Therefore, we can solve for V0 by substituting t = total time in the air and y = 0 into the kinematic equation:
0 = h0 + V0*t - (1/2)*g*t^2
0 = 2 + V0*t - (1/2)*(-9.8)*t^2
0 = 2 + V0*t + 4.9*t^2
Now we can use the value of V0 to solve for t, and we will get the total time the ball is in the air.
To find the time the ball is in the air, we need to use the kinematic equation for vertical motion:
h = ut + 0.5gt^2
Where:
- h is the height (2 m in this case)
- u is the initial vertical velocity (which for this case is 0 since the ball is thrown straight up)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
Since the initial vertical velocity is 0, the equation simplifies to:
h = 0.5gt^2
Rearranging the equation to solve for time:
t^2 = (2h) / g
t = sqrt((2h) / g)
Now, let's plug in the values:
t = sqrt((2 * 2) / 9.8)
t = sqrt(4/9.8)
t ≈ 0.90 s
So, the ball is in the air for approximately 0.90 seconds.