Answer :
assuming that the ball is launch from the ground. the initial velocity would be equal to its final velocity when it reach the ground. this is because the acceleration due to gravity is working on the ball, when it goes up decelerates up to zero, then falls back then acceleartes again until it reach the groun, where the velocity is 150 m/s
Step-by-step explanation:
It is given that,
A baseball is hit straight up at an initial velocity of 150 m/s. Firstly calculating the maximum height of the balloon. It can be calculated suing third equation of motion as :
[tex]v^2-u^2=-2gh[/tex]
At maximum height v = 0
[tex]h=\dfrac{u^2}{2g}[/tex]
[tex]h=\dfrac{(150\ m/s)^2}{2\times 9.8\ m/s^2}[/tex]
h = 1147.95 m
or
h = 1148 m
We have to find the baseball's speed when it hits the ground. At this point, its initial velocity is 0 and when its hits the ground its velocity is v'. Again using third equation of motion and finding the value of v' as :
[tex]v'^2-u^2=2gh[/tex]
[tex]v'=\sqrt{2gh}[/tex]
[tex]v'=\sqrt{2\times 9.8\ m/s^2\times 1148\ m}[/tex]
v'= 150 m/s
When the baseball hits the ground, its speed is equal to 150 m/s. Hence, this is the required solution.