Question

❤✨HOLLA✨❤

✨✨Physics Genius answer it ✨✨

● A Baseball is popped straight up into the air and Has a hang time of 6.25s . Determine the height to which ball rises before it reaches its peak . (Hint the time to rise to the peak is one - half the total hang time )

✔I will mark you a Brainliest ✔​

Answer 1

Answer:

At the maximum height of the ball, the velocity will be zero. This will also take half the total time, 3.125 s. 

Now use the equation... 

Δd = vf*t - 1/2at^2 

By setting the final time at the moment of the apex of the height, vf becomes 0. 

So the equation will become... 

Δd = - 1/2at^2 

The equation now resembles the change in height from the ground to the maximum height 

Δd = - 1/2at^2 = - 1/2 (-9.8 m/s^2) (3.125 s)^2 = 47.8 m

Answer 2

${\mathfrak{\red{\underline{\underline{Answer:-}}}}}$

=> h = 47.851 m

${\mathfrak{\red{\underline{\underline{Explanation:-}}}}}$

Here time of flight is, t = 6.25 s

Acceleration due to gravity, g = -9.8 m/s²

Time to reach the maximum height = t/2 = 3.125 s

At the maximum height ‘h’ the velocity becomes zero. Let the initial velocity be ‘u’.

So, 0 = u –(9.8)(3.125)

=> u = 30.625 m/s

Now,

h = u(t/2) + ½ g(t/2)²

=> h = (30.625)(3.125) – (0.5)(9.8)(3.125)²

=> h = 47.851 m

Hope it helps!!