A ball dropped from a sixth floor window accelerates at 10 m/s
After 2 seconds it hits the
ground. It bounces back with a
velocity of
-15 m/s.
What is the ball’s velocity
when it hits the ground?
What is the change in velocity of the ball when it bounces?
The ball is in contact with the ground for 0.2 s when it bounces. What is
its mean acceleration while it is
in contact with the ground?
think. The velocity at impact had to be 15m/s if the bounce was elastic.
change velocity=(vf-Vi)=-15-15=-30m/s
Acceleration= changevelociyt/time
To solve this problem, we need to break it down into steps:
Step 1: Find the velocity when the ball hits the ground.
Given:
Acceleration (a) = 10 m/s^2
Time (t) = 2 s
Initial velocity (u) = 0 (since the ball was dropped)
Using the formula: v = u + at
v = 0 + 10 * 2
v = 20 m/s
Therefore, the ball's velocity when it hits the ground is 20 m/s.
Step 2: Find the change in velocity of the ball when it bounces.
Given:
Initial velocity after bouncing (u') = -15 m/s
Final velocity after bouncing (v') = 20 m/s
Since the ball bounces, the direction of the velocity changes. So, we'll consider the magnitudes.
Using the formula: Δv = v' - u'
Δv = 20 - (-15)
Δv = 35 m/s
Therefore, the change in velocity of the ball when it bounces is 35 m/s.
Step 3: Find the mean acceleration while the ball is in contact with the ground.
Given:
Time of contact with the ground (t') = 0.2 s
Change in velocity (Δv) = 35 m/s
Using the formula: a' = Δv / t'
a' = 35 / 0.2
a' = 175 m/s^2
Therefore, the mean acceleration while the ball is in contact with the ground is 175 m/s^2.
To find the ball's velocity when it hits the ground, we can use the equation: velocity = initial velocity + acceleration * time.
The ball was dropped, so its initial velocity is 0 m/s.
The acceleration is given as -10 m/s (assuming downwards is the positive direction).
The time it takes to hit the ground is given as 2 seconds.
Using the formula, we can calculate the velocity when it hits the ground:
velocity = 0 + (-10) * 2
velocity = -20 m/s
Therefore, the ball's velocity when it hits the ground is -20 m/s.
To find the change in velocity of the ball when it bounces, we can subtract the final velocity before the bounce from the initial velocity after the bounce.
The final velocity before the bounce is -15 m/s (assuming upwards is the negative direction).
The initial velocity after the bounce is -15 m/s (assuming downwards is the positive direction).
change in velocity = initial velocity - final velocity
change in velocity = (-15) - (-15)
change in velocity = 0 m/s
Therefore, the change in velocity of the ball when it bounces is 0 m/s.
To find the mean acceleration while the ball is in contact with the ground, we can use the equation: acceleration = change in velocity / time.
The change in velocity is 0 m/s (as calculated above).
The time the ball is in contact with the ground is given as 0.2 seconds.
Using the formula, we can calculate the mean acceleration:
acceleration = 0 / 0.2
acceleration = 0 m/s^2
Therefore, the mean acceleration of the ball while it is in contact with the ground is 0 m/s^2.