Answer:
a) The velocity of the ball before it hits the floor is -6.3 m/s
b) The velocity of the ball after it hits the floor is 3.1 m/s
c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.
Explanation:
First, let´s calcualte how much time it takes the ball to hit the floor:
The equation for the position of the ball is:
y = y0 + v0 * t + 1/2 g * t²
Where:
y = position at time t
y0 = initial position
v0 = initial velocity
t = time
g = acceleration due to gravity
We take the ground as the origin of the reference system.
a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:
0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s² * t²
-2.0 m = -4.9 m/s² * t²
t² = -2.0 m / - 4.9 m/s²
t = 0.64 s
The equation for the velocity of a falling object is:
v = v0 + g * t where "v" is the velocity
since v0= 0:
v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s
b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:
v = v0 + g * t
0 = v0 + g * t
-v0/g = t
now we replace t in the equation for position, since we know that the maximum height is 1.5 m:
y = y0 + v0 * t + 1/2* g * t² y = 1.5 m y0 = 0 m t = -v0/g
1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²
1.5 m = - v0²/g - 1/2 * v0²/g
1.5 m = -3/2 v0²/g
1.5 m * (-2/3) * g = v0²
1.5 m * (-2/3) * (-9.8 m/s²) = v0²
v0 = 3.1 m/s
c) The average acceleration will be:
a = final velocity - initial velocity / time
a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²
the direction of the acceleration is upward perpendicular to the ground.
The vector average acceleration will be:
a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)
