A ball is dropped from rest from the 4th story of a building, 14 meters off of the ground.
How much time passes before the ball hits the ground?
What is the ball’s velocity (speed AND direction) the split-second before it hits the ground? This answer is not zero - it hasn’t hit the ground yet!
Would I use the equation Vf=Vo+at
well, eventually maybe
h = Hi + Vi t + .5 a t^2 and a = -9.81
0 = 14 + 0 - 4.9 t^2
t^2 = 14/4.9
t = 1.69 seconds
v = Vi + a t
v = 0 -9.81(1.69)
v = - 16.6 m/s
The thing is, the only formulas my teacher gave me are these
Vf=Vo+at
Xf=Xo+Vot + 1/2at^2
Vf^2=Vo^2+2a(Xf-Xo)
So I dont want to use formulas I havent been taught but thank you!! But can you solve this problem with any of the above formulas
To find the time it takes for the ball to hit the ground, you can use the kinematic equation:
h = (1/2)gt^2
Where:
h is the height (distance) the ball has fallen (14 meters)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time it takes for the ball to fall
Rearranging the equation to solve for t, we get:
t = sqrt(2h / g)
Plugging in the values given, we have:
t = sqrt(2 * 14 / 9.8)
t ≈ 1.92 seconds
Therefore, it takes approximately 1.92 seconds for the ball to hit the ground.
Now, to find the velocity of the ball just before it hits the ground, we can use another kinematic equation:
Vf = Vo + at
Where:
Vf is the final velocity (which we'll find)
Vo is the initial velocity (which is 0 since the ball is dropped from rest)
a is the acceleration (which is the acceleration due to gravity, -9.8 m/s^2)
t is the time calculated earlier (1.92 seconds)
Plugging in the values, we have:
Vf = 0 + (-9.8) * 1.92
Vf ≈ -18.816 m/s
The negative sign indicates that the velocity is directed downwards. Therefore, the ball's velocity just before hitting the ground is approximately 18.816 m/s downwards.