Ayoung child throw a ball upwards with an initial velocity of 20 feet/second. the height of the ball (in feets) after t seconds is given by the function h(t)=16^2+20t+3. what is the maximum height the ball

If this is a parabolic motion equation, then it is a negative parabola, which looks like a hill (instead of a positive parabola that opens like a cup). Your equation would be h(t)= -16t^2 + 20t +3. That's the equation for an initial velocity of 20 ft/s thrown from an initial height of 3 ft. And the -16t^2 is the antiderivative of the gravitational pull. Anyway, if you're looking for the maximum height and you don't know calculus, then you have to complete the square to get this into vertex form. The vertex will be the highest point on the graph, which is consequently also the max height of the ball. When you do this, you get a vertex of (5/8, 9.25). The 9.25 is the max height of the ball.

11/2.2 = 5 inches for the length 8.5/2.2= 3.86 inches for the width