If a ball is dropped from a height of 100.0 m, what is it's height above the ground (in meters) after 3.0 s?
If a ball is dropped from a height of 100.0 m, what is it's velocity (in m/s) after 3.00 s? You may treat downward velocity as positive.
v = 9.8t
s = 100 - 4.9t^2
To find the height of the ball above the ground after 3.0 s, you can use the equation of motion for freefall:
h = h0 + v0*t + (1/2)*a*t^2
where:
h = final height above the ground
h0 = initial height (given as 100.0 m)
v0 = initial velocity (which is 0 m/s since the ball is dropped)
t = time (given as 3.0 s)
a = acceleration due to gravity (~9.8 m/s^2)
Substituting the given values into the equation:
h = 100.0 m + 0 m/s * 3.0 s + (1/2) * 9.8 m/s^2 * (3.0 s)^2
h = 100.0 m + 0 + 44.1 m
h ≈ 144.1 m
Therefore, the ball's height above the ground after 3.0 s is approximately 144.1 meters.
To find the velocity of the ball after 3.0 s, you can use the equation for velocity during freefall:
v = v0 + a*t
Again, substituting the given values into the equation:
v = 0 m/s + 9.8 m/s^2 * 3.0 s
v = 0 + 29.4 m/s
v = 29.4 m/s
Therefore, the ball's velocity after 3.0 s is 29.4 m/s.