Question

13. A ball is thrown downward from the top of a building with an initial speed of 25 m/s. It
strikes the ground after 20 s. How high is the building, assuming negligible air resistance?

Answer 1

To find the height of the building, we can use the formula:

h = h0 + v0t + (1/2)gt^2,

where h is the final height, h0 is the initial height (height of the building), v0 is the initial velocity (throwing downwards), t is the time it takes for the ball to hit the ground, and g is the acceleration due to gravity (approximately 9.8 m/s^2).

We are given:
v0 = 25 m/s (initial velocity)
t = 20 s (time taken to hit the ground)
g = 9.8 m/s^2 (acceleration due to gravity)

Now, let's plug in the values into the formula:

h = h0 + v0t + (1/2)gt^2
h = h0 + (25 m/s)(20 s) + (1/2)(9.8 m/s^2)(20 s)^2
h = h0 + 500 m + (1/2)(9.8 m/s^2)(400 s^2)
h = h0 + 500 m + 1,960 m
h = h0 + 2,460 m

Since the ball starts from the top of the building, the initial height (h0) is equal to the height of the building. Therefore, we can simplify the equation to:

h = h0 + 2,460 m

So, the height of the building is equal to 2,460 meters.