The time at which both balls will be at the same height is 0.6 seconds.
The initial speed at which the ball is thrown upward =
15.2 m/s
Height of the building from where another ball is dropped = 19 m
Let the distance travelled by the first ball is d.
d = (10 m - h)
The height of the ball that is falling from a 10 m high building is,
[tex]y _{f} = y _{i} + v _{i} - \frac{1}{2} gt ^{2} [/tex]
[tex](-10 -h) = \frac{1}{2} gt^{2} [/tex]
[tex]h = 10 - \frac{1}{2} gt ^{2} [/tex]
The height of the ball that is thrown upward with 15.2 m/s is,
[tex]y _{f} = y _{i} + v _{i} - \frac{1}{2} gt ^{2} [/tex]
[tex]h = (15.2)t- \frac{1}{2} gt ^{2} [/tex]
The time at which both balls will be at the same height is,
[tex] (15.2)t- \frac{1}{2} gt ^{2} = 10 - \frac{1}{2} gt ^{2}[/tex]
[tex]t = \frac{10}{15.2} [/tex]
t = 0.6 s
Therefore, the time at which both balls will be at the same height is 0.6 seconds.
To know more about initial speed, refer to the below link:
https://brainly.com/question/13033040
#SPJ4
