I have solved the problem but am not sure it it is right. I would appreciate if you would check to see if my answer is right.
The height of the ball thrown upwards from the top of a 20 foot hill is given by h(t)= -16t^2+90t+20 ft (where t is measured in seconds). how high above the ground is the ball at its highest point.
I took the derivative and got -32t+90=0
-2(16t-45)=0
t=45/16
h(45/16)=-16(45/16)^2+90(45/16)+20
h(45/16)=146.6 ft
Looks good to me.
To find the highest point of a ball's trajectory, we need to find the vertex of the parabolic function h(t) = -16t^2 + 90t + 20.
To do this, you correctly took the derivative of h(t) with respect to t and set it equal to zero:
h'(t) = -32t + 90 = 0
Simplifying this equation, we get:
-32t = -90
t = -90 / -32
t = 45/16
So, you correctly found that the time at the highest point is t = 45/16 seconds.
To find the height at the highest point, we substitute this value back into the original function:
h(45/16) = -16(45/16)^2 + 90(45/16) + 20
h(45/16) = -16(2025/256) + 90(45/16) + 20
h(45/16) ≈ 146.6 ft
Based on your calculations, it appears that the height of the ball at its highest point is approximately 146.6 feet.