A ball is dropped from a high building. Using the approximate value of g = 10 m/s2, find the change in velocity between the fourth and seventh second of its flight

find v(7) : v(7)=10*7=70

find v(4) : v(4)=10*4=40

calculate the change.

30 is the change

30m/s

Why did the ball go to a therapist after being dropped from a high building? Because it had a sudden change in velocity and needed some counseling!

But to get serious, let's calculate the change in velocity. In the first second, the velocity of the ball would be 10 m/s. After the second second, it would be 20 m/s (assuming no air resistance).

So, between the fourth and seventh second, the velocity would increase by 30 m/s. The ball would speed up! That's quite the change, isn't it?

To find the change in velocity between the fourth and seventh second of the ball's flight, we need to calculate the velocity at both of these time points and then subtract the initial velocity from the final velocity. Let's break it down into steps:

Step 1: Determine the initial velocity.
Since the ball is dropped from rest, the initial velocity (u) is 0 m/s.

Step 2: Calculate the velocity at the fourth second.
We can use the formula for velocity (v) in free fall: v = u + gt, where g is the acceleration due to gravity and t is the time. In this case, g = 10 m/s^2 and t = 4 seconds. Plugging in these values, we get:
v = 0 + (10 * 4) = 40 m/s.

Step 3: Calculate the velocity at the seventh second.
Using the same formula, we now have t = 7 seconds:
v = 0 + (10 * 7) = 70 m/s.

Step 4: Find the change in velocity.
To get the change in velocity, we subtract the initial velocity from the final velocity:
Δv = v_final - v_initial = 70 m/s - 40 m/s = 30 m/s.

Therefore, the change in velocity between the fourth and seventh second of the ball's flight is 30 m/s.