A ball is battled straight up and returns to the level of the batter after 6.5s determine how high the ball went
it is at the peak at 6.5/2 seconds...
and the velocity at the top is zero..
so consider it going down...
hf=hi+vi*t-1/2 g t^2
0=hi+0*t-1/2 g t^2
you know g, solve for hi
To determine how high the ball went, we need to understand the motion of the ball and use some basic physics equations.
When the ball is thrown straight up, it follows a vertical path, experiencing only the force of gravity pulling it downwards. The ball reaches its maximum height when its vertical velocity becomes zero.
We can use the following equation to determine the height:
h = (v^2) / (2g)
Where:
h = height
v = initial vertical velocity
g = acceleration due to gravity
In this case, since the ball returns to the level of the batter after 6.5 seconds, we know that it took a total of 6.5 seconds to complete the entire flight, going up and coming back down.
The time taken to reach maximum height is half of the total time, as the time taken to rise is equal to the time taken to fall. So, the time to reach maximum height is 6.5/2 = 3.25 seconds.
Now, we need to find the initial vertical velocity (v). At the maximum height, the vertical velocity is zero. We can use the equation:
v = u + gt
Where:
v = final vertical velocity (zero at maximum height)
u = initial vertical velocity
g = acceleration due to gravity
t = time taken to reach maximum height (3.25 seconds)
Rearranging the equation:
u = -gt
Substituting the values:
u = -(9.8 m/s^2) * (3.25 s)
u = -31.85 m/s
Now, we have the initial vertical velocity (u). We can substitute this value into the equation for height:
h = (v^2) / (2g)
h = (0^2) / (2 * -9.8 m/s^2)
h = 0 m
Therefore, the ball did not reach any height. It returned to the level of the batter without going higher.