A little girl is standing on a cliff and throws a ball straight down at a speed of 5m/s towards the ground below. How far has it fallen 2.0 seconds after it leaves her hand?
starting at height 0, its height at time t is
-5t - 4.9t^2
If we fudge a bit and approximate gravity with 5 instead of 4.9, then we have
-5t-5t^2 = -5t(1+t)
at t=2, that's -30
To determine how far the ball has fallen after 2.0 seconds, we need to use the equation for the distance fallen by an object in free fall. The equation is given by:
d = (1/2)gt^2
where:
d = distance fallen
g = acceleration due to gravity (approximately 9.8 m/s^2 near the surface of the Earth)
t = time in seconds
In this case, the ball is thrown downwards with an initial speed of 5 m/s, so its initial velocity is -5 m/s (negative because it is in the downward direction). The equation needs to be modified to take the initial velocity into account:
d = v0t + (1/2)gt^2
where:
v0 = initial velocity
Substituting the given values into the equation, we have:
d = -5 * 2 + (1/2) * 9.8 * (2^2)
Simplifying the equation further:
d = -10 + (1/2) * 9.8 * 4
d = -10 + 19.6
d = 9.6 meters
Therefore, the ball has fallen 9.6 meters after 2.0 seconds.