Question
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
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
Answers
Steve
Looks good to me.