00:01
Hi, the function is given as h of t is equal to minus 16 t squared plus 56t and then we have plus 32 right, given as here plus 32 right this model the ball site we have above the ground this model the ball site above the ground basically in 50 seconds after it was thrown.
00:29
If i made a make a graph of this here we have t, here we have h of t right so we have t and h of t right.
00:37
Here initial point is 32 here, there's 32 here, and the coefficient of t squared is negative.
00:43
So the parabola will open downwards, right, like this.
00:46
Here we have.
00:47
So at this point here we have the maximum height.
00:52
Maxim height is there, right? at maximum height we have x coordinate is minus b over 2a.
00:58
And accordingly we can find the y coordinate to find the maximum height.
01:01
So if you compare this equation with a x square plus bx plus c, we get as minus 16.
01:10
B as 56 and c is 32.
01:14
So we'll have x is equal to minus b by 2a minus 56 over 2 times minus 16.
01:20
So minus 56 over we have 2 times minus 16.
01:25
That is given as 1 .75.
01:26
That will be 1 .75 seconds, right? that is g basically, right? that is given as g basically 1 .75 seconds, right? now we need to find the y value here.
01:37
So h of 1 .75, h of, 1 .75 that is given as minus 16 1 .75 square we have 1 .75 square this we have plus 56 times 1 .75 this we have right plus we have 32 right plus we have 32 here so minus 16 times 1 .75 square plus 56 times 1 .75 and then we have plus 32 right we have here.
02:19
Now, if you just solve this, minus 16 times 1 .75 square plus 56 times 1 .75 plus 32.
02:26
That is giving 81...