A ball dropped from a sixth floor window accelerates at 10m/s squared. After 2 seconds it hits the ground. It bounces back with a velocity of -15m/s squared.
A. What is the balls velocity when it hits the ground?
b. The ball is in contact with the ground for 0.02 s when it bounces. What is its mean acceleration while it is in contact with the ground?
two problems here:
1. velocity is not in m/s^2
2. It should bounce back with an upward (positive) velocity.
sorry
A ball dropped from a sixth floor window accelerates at 10m/s squared. After 2 seconds it hits the ground. It bounces back with a velocity of -15m/s.
this is what the worksheet says, that is why im soo confused
see the problem
http://www.jiskha.com/display.cgi?id=1380223491
Steve said it should bounce back at 15, but that presumes upward as + direction. One can make it the - direction, but that is seldom done. The direction signs are arbritary.
To answer these questions, we need to use the principles of motion and acceleration.
A. What is the ball's velocity when it hits the ground?
We can use the second equation of motion to calculate the ball's final velocity when it hits the ground:
v = u + at
Where:
v = Final velocity (unknown)
u = Initial velocity (0 m/s, as the ball was dropped)
a = Acceleration (10 m/s^2, as given in the question)
t = Time (2 seconds, as given in the question)
Plugging in the values, we get:
v = 0 + (10 m/s^2)(2 s)
v = 20 m/s
Therefore, the ball's velocity when it hits the ground is 20 m/s.
B. What is its mean acceleration while it is in contact with the ground?
Mean acceleration can be calculated using the equation:
Acceleration (mean) = Change in velocity / Time taken
The change in velocity can be determined by subtracting the initial velocity from the final velocity:
Change in velocity = Final velocity - Initial velocity
Given:
Initial velocity (just before the bounce) = 20 m/s (downwards)
Final velocity (just after the bounce) = -15 m/s (upwards)
Time taken for the bounce = 0.02 seconds
Change in velocity = (-15 m/s) - (20 m/s)
Change in velocity = -35 m/s
Now, we can calculate the mean acceleration:
Acceleration (mean) = (-35 m/s) / (0.02 s)
Acceleration (mean) = -1750 m/s^2
So, the ball's mean acceleration while in contact with the ground is -1750 m/s^2. Note that this acceleration is negative because it is opposite to the direction of the ball's initial motion.