Variations

Question

At 9:45 Margie threw a ball upwards while standing on a platform 65 ft above the ground. The trajectory after t seconds follows the equation: -0.8t^2+56t+65.

A)What will be the maximum height of the ball?

b)How long will it take the ball to reach is maximum height?

 

I came to the answer of 3005 ft for A

and 35 seconds for B.

Is this right?

Answer 1

Let us assume;

- y is the height above the roof top, and

- h is the height above the ground

 

according to the trajectory equation;

 

y = -0.8t^2+56t+65

 

to find out maximum y;

 

dy/dt = 0 = -1.6t + 56

this gives t = 35 seconds

 

Therefore, the trajectory can reach its maximum y within 35 seconds

 

Let us put 35 seconds in the trajectory equation, so that we will get ymax

 

ymax = -0.8(35)^2 + 56(35) +56

         = 1036 ft

 

Remember, Margie is already at the top of platform 65 ft above the ground!

 

Therefore, the maximum height of the trajectory above the ground is:

 

hmax = ymax + 65

         = 1036 + 65

         = 1, 101 ft