Question

(1 point) Basil drops a baseball from the top of a 88 m tall building.

(a) Determine the speed of the ball 2's after it was dropped.
m/s
(b) Determine how far the ball falls after 2 s.
m
(c) Determine when the ball reaches the ground.
I
s

Answer 1

a) The magnitude of the speed of the ball at 2s is 19.62 m/s.

b) The ball has traveled 19.62 m after 2 seconds

c) The ball reaches the ground in 4.24 s.

Step-by-step explanation:

a) The speed of the ball at 2 s after it was dropped is:

`v_(f) = v_(0) - gt`

Where:

`v_(f)`: is the final speed =?

`v_(0)`: is the initial speed = 0 (it is dropped)

g: is the gravity = 9.81 m/s²

t: is the time = 2 s

`v_(f) = -9.81 m/s^(2)*2 s = -19.62 m/s`

Then, the speed of the ball at 2s is -19.62 m/s. The minus sign is because the speed is in the negative direction (down).

b) The height at which is the ball after 2 seconds is:

`y_(f) = y_(0) + v_(0)t - (1)/(2)gt^(2)`

Taking y₀ = 0 we have:

`y_(f) = 0 - (1)/(2)*9.81 m/s^(2)*(2 s)^(2) = -19.62 m`

The ball has traveled 19.62 m after 2 seconds. The minus sign is because the height is in the negative direction.

c) The time at which the ball reaches the ground is:

`y_(f) = y_(0) + v_(0)t - (1)/(2)gt^(2)`

`t = \sqrt{(88 m)/((1)/(2)*9.81 m/s^(2))} = 4.24 s`

Therefore, the ball reaches the ground in 4.24 s.

I hope it helps you!