Projectile motion of baseball

$\Delta y$ , which is the vertical displacement, or height, can be expressed by the following equation: $\Delta y = v_{i,y}t - \frac{1}{2}gt^2$ .

By letting $\Delta y = 25$ , $t = 1$ , and $g = 10$ , one can solve for $v_{i,y}$ :

$v_{i,y} = \Delta y + \frac{1}{2}gt^2$

$v_{i,y} = 30$

At the peak of its trajectory, we know that the vertical component of its velocity is zero, that is, $v_{f,y} = 0$ . Using the formula $v_{f,y} = v_{i,y} - gt$ :

$t = \frac{v_{f,y} - v{i,y}}{-g}$

$t = \boxed{3}$

x=ut+ a(t*t)/2

as t=1 ,a=-10 ==> u=30

v=u+at

at highest point v=0

==> 0=30-10(t)

==> t=3

If the projectile went 25m in 1 second, then the average speed was 25m/s. Therefore, the initital speed was 30m/s and the final was 20m/s averaging 25m/s over that 1 second interval. At 1 second later it will lose another 10m/s due to gravity and be traveling at 10m/s. Finally it has 1 second more to get to 0m/s. This is 3 seconds. In summary at 0s it is moving 30m/s, at 1s it is moving 20m/s, t 2s it is moving 10m/s and at 3s it is moving 0m/s,

because projection speed is 30 m/sec