Answer:
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.
Explanation:
a) The speed of the ball at 2 s after it was dropped is:
[tex] v_{f} = v_{0} - gt [/tex]
Where:
[tex]v_{f}[/tex]: is the final speed =?
[tex]v_{0}[/tex]: is the initial speed = 0 (it is dropped)
g: is the gravity = 9.81 m/s²
t: is the time = 2 s
[tex] v_{f} = -9.81 m/s^{2}*2 s = -19.62 m/s [/tex]
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:
[tex] y_{f} = y_{0} + v_{0}t - \frac{1}{2}gt^{2} [/tex]
Taking y₀ = 0 we have:
[tex] y_{f} = 0 - \frac{1}{2}*9.81 m/s^{2}*(2 s)^{2} = -19.62 m [/tex]
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:
[tex] y_{f} = y_{0} + v_{0}t - \frac{1}{2}gt^{2} [/tex]
[tex] t = \sqrt{\frac{88 m}{\frac{1}{2}*9.81 m/s^{2}}} = 4.24 s [/tex]
Therefore, the ball reaches the ground in 4.24 s.
I hope it helps you!
