a student drops a ball from a window 3.5m above the sidewalk. the ball accelerates at 9.80m/s^2. How fast is it moving when it hits the sidewalk.
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*3.5 = 68.6,
Vf = 8.28m/s.
Vf^2= Vi^2 + 2ad
Vf^2= 2(-9.80)(3.5)
Vf^2= -68.6
(square root Vf^2 and drop the - on 68.6 and square root it)
Vf= 8.3 m/s down
(answer is rounded up to 2 sig figs)
Well, since the ball was dropped from a window, we can safely assume it wasn't having a "falling out" with anyone. As for its speed when it hits the sidewalk, we can calculate that using the good ol' laws of physics.
We'll use the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Since the ball was dropped, its initial velocity is 0 m/s (it wasn't doing any fancy tricks up there). The acceleration is given as 9.80 m/s² (gravity pulling it down). And the distance traveled is 3.5 meters (from the window to the sidewalk).
Now we just need to rearrange the equation to solve for v:
v² = 0² + 2 * 9.80 * 3.5
That gives us v² = 68.6
So, when we carefully extract the square root, we find that v is approximately equal to √68.6 m/s.
But let's not forget our clown physics where we always have to convert meters to laughter per second. After some silly calculations, the answer is approximately v = 8.29 m/s.
So, when the ball hits the sidewalk, it'll be moving at approximately 8.29 meters per second. Just make sure to catch it, or else it'll have a "ground-breaking" performance!
To find the speed at which the ball is moving when it hits the sidewalk, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is what we want to find)
u = initial velocity (0 m/s, since the ball was dropped)
a = acceleration (9.80 m/s^2)
s = distance (3.5 m)
Step 1: Substitute the given values into the equation.
v^2 = 0^2 + 2(9.80)(3.5)
Step 2: Simplify the equation.
v^2 = 2(9.80)(3.5)
Step 3: Calculate the right-hand side of the equation.
v^2 = 68.6
Step 4: Take the square root of both sides of the equation.
v = √68.6
Step 5: Calculate the square root.
v ≈ 8.29 m/s
Therefore, the ball is moving at approximately 8.29 m/s when it hits the sidewalk.
To determine the speed of the ball when it hits the sidewalk, we can use the equations of motion in physics.
The first equation we can use is the equation of motion for displacement:
d = vit + 0.5at^2
Where:
d = displacement (in this case, the distance the ball falls, which is 3.5m)
vi = initial velocity (since the ball is dropped, the initial velocity is 0)
a = acceleration (given as -9.80m/s^2, negative because the acceleration is directed downwards)
t = time
We can rearrange the equation to solve for time (t):
t = √(2d / a)
Plugging in the values given:
d = 3.5m
a = -9.80m/s^2
t = √(2 * 3.5m / -9.80m/s^2)
Calculate the value inside the square root first:
2 * 3.5m = 7m
t = √(7m / -9.80m/s^2)
Now, we can calculate the square root:
t ≈ 0.94 seconds (rounded to 2 decimal places)
Now that we know the time it takes for the ball to reach the ground, we can find its final velocity using:
vf = vi + at
Where:
vi = initial velocity (0m/s)
a = acceleration (-9.80m/s^2)
vf = final velocity (what we need to find)
Plugging in the values:
vi = 0m/s
a = -9.80m/s^2
t ≈ 0.94 seconds
vf = 0m/s + (-9.80m/s^2) * 0.94s
Calculate the multiplication first:
(-9.80m/s^2) * 0.94s ≈ -9.212m/s
vf ≈ 0m/s + (-9.212m/s) ≈ -9.21m/s
The negative sign indicates that the velocity is directed downwards. Therefore, the speed of the ball when it hits the sidewalk is approximately 9.21m/s.