A ball is thrown straight up with an initial speed of 60 m/s. How high does it go? Assume the accelera-
tion of gravity is 10 m/s22
2)How long is it in the air?
1. Vf^2 + 2ad = 0,
(60)^2 + 2 * (-10)d = 0,
3600 - 20d = 0,
-20d = -3600,
d = 180 m upward.
2. d = Vo*t + 0.5at^2 = 180 m upward,
60t + 0.5*(-10)t^2 = 180,
60t - 5t^2 = 180,
Rearrange terms:
-5t^2 + 60t = 180,
Divide both sides by -5:
t^2 - 12t = - 36,
t^2 - 12t + 36 = 0,
Factor and get:
(t - 6) (t - 6) = 0,
t - 6 = 0,
t = 6 s = t(up).
d = 0.5at^2 = 180 m downward,
0.5 * 10t^2 = 180,
5t^2 = 180,
t^2 = 36,
t = 6 s downward.
t(up) + t(down) = 6 + 6 = 12 s in air.
To solve this problem, we can use the equations of motion for an object in free fall under the acceleration due to gravity.
1) How high does the ball go?
To find the maximum height reached by the ball, we can use the following equation:
h = (v^2 - u^2) / (2g)
Where:
- h is the maximum height,
- v is the final velocity (which is zero at the highest point),
- u is the initial velocity,
- g is the acceleration due to gravity.
Given:
- u = 60 m/s
- g = 10 m/s^2
Substituting these values into the equation, we have:
h = (0 - 60^2) / (2 * 10)
Simplifying the equation, we get:
h = (-3600) / 20
h = -180 m
The negative sign indicates that the ball is at a height of 180 m above its initial position.
2) How long is it in the air?
To find the time the ball is in the air, we can use the following equation:
t = (v - u) / g
Where:
- t is the time of flight,
- v is the final velocity (which is zero at the highest point),
- u is the initial velocity,
- g is the acceleration due to gravity.
Using the given values:
t = (0 - 60) / 10
t = -6 s
Again, the negative sign indicates that the ball is in the air for 6 seconds.