A ball is thrown straight up with an initial
speed of 60 m/s.
How high does it go? Assume the acceleration of gravity is 10 m/s
2
.
Answer in units of m
010 (part 2 of 2) 10.0 points
How long is it in the air?
Answer in units of s
h=v²/2g
To find the height the ball goes, we can use the equation for vertical motion:
hf = hi + vi*t - (1/2)g*t^2
Here, hf is the final height, hi is the initial height (which is 0 in this case), vi is the initial velocity (60 m/s), g is the acceleration due to gravity (-10 m/s^2), and t is the time the ball is in the air.
Since the ball is thrown straight up, the final height will be equal to the initial height (0). So we can rearrange the equation to solve for t:
0 = 0 + 60*t - (1/2)*(-10)*t^2
Simplifying the equation gives:
5t^2 - 60t = 0
We can factor out t:
t * (5t - 60) = 0
So either t = 0 or t = 60/5 = 12 s.
Since we are interested in the time the ball is in the air, t = 12 s is the correct answer.
Therefore, the ball is in the air for 12 seconds.