Answer:
a) The ball's velocity just prior to hitting the ground measures -6.3 m/s
b) The ball's velocity right after bouncing off the ground registers at 3.1 m/s
c) The average acceleration's magnitude is 470 m/s², and its direction is upward, forming a 90º angle with the ground.
Explanation:
To begin, let’s assess the time it takes for the ball to reach the floor:
The equation outlining the ball's position is:
y = y0 + v0 * t + 1/2 g * t²
Where:
y = position at given time t
y0 = initial position
v0 = initial velocity
t = time
g = acceleration triggered by gravity
We establish the ground as the reference origin.
a) Since the ball is released rather than thrown, the initial velocity v0 is 0. The direction of acceleration is downward, directed towards the origin; thus, “g” is treated as negative. When the ball contacts the ground, its position will be 0. Therefore:
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 motion equation for a falling body is:
v = v0 + g * t where "v" denotes the velocity
Since v0= 0:
v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s
b) The pebble's speed reaches 0 during its maximum height. To find the time taken for the pebble to achieve that height, we can use the velocity equation and then substitute that time in the position equation to derive the initial velocity:
v = v0 + g * t
0 = v0 + g * t
-v0/g = t
Replacing t in the position equation, knowing 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 can be determined by:
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)
