(a) How long does it take for a golf ball to fall from rest for a distance of 12.5 m?in seconds
(b) How far would the ball fall in twice that time?in meters
h=Vi*t +1/2 g t^2 solve for t.
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To answer these questions, we can use the equations of motion under constant acceleration. The first equation we'll use is:
d = ut + (1/2)at^2
where:
d is the distance traveled,
u is the initial velocity (which is 0 since the ball is at rest),
t is the time taken,
and a is the acceleration due to gravity (which is approximately 9.8 m/s^2).
(a) How long does it take for a golf ball to fall from rest for a distance of 12.5 m?
Using the equation above, we can rearrange it to solve for t:
t = sqrt((2d) / a)
Plugging in the values, we get:
t = sqrt((2 * 12.5) / 9.8)
Simplifying, we have:
t = sqrt(25 / 9.8)
Calculating this, t ≈ 1.59 seconds.
Therefore, it takes approximately 1.59 seconds for the golf ball to fall from rest for a distance of 12.5 meters.
(b) How far would the ball fall in twice that time?
Since we know the time it takes for the ball to fall (1.59 seconds), we can find the distance traveled using the same equation:
d = ut + (1/2)at^2
But this time, the initial velocity u is not zero, as the ball has been falling for some time. So, we need to determine the initial velocity in this case:
u = gt
Plugging in the values:
u = 9.8 * 1.59
Calculating this, u ≈ 15.58 m/s.
Now, we can use the equation to calculate the distance traveled in twice the time:
d = ut + (1/2)at^2
d = (15.58 * 2) + (1/2) * 9.8 * (1.59^2)
Calculating this, d ≈ 24.7 meters.
Therefore, the ball would fall approximately 24.7 meters in twice the time it took to fall 12.5 meters.