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
-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
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.