Suppose that a baseball is tossed up into the air at an initial velocity 1818​​\text{ m/s} m/s​. The height of the baseball at time tt​ in s

Question

Suppose that a baseball is tossed up into the air at an initial velocity 1818​​\text{ m/s} m/s​. The height of the baseball at time tt​ in seconds is given by h\mathopen{}\mathclose{\left(t\right)}=18 t-4.9 t^{2}h(t)=18t−4.9t 2 ​​​​​ (in meters). a) What is the average velocity for \left[1, 1.5\right][1,1.5]​?

in progress 0
11 hours 2021-07-21T21:07:32+00:00 1 Answers 0 views 0

13.1 m/s

Explanation:

Given that a baseball is tossed up into the air at an initial velocity 18 m/s​. The height of the baseball at time t​ in seconds is given by h(t) = 18t−4.9t 2 ​​​​​ (in meters).

a) What is the average velocity for [1,1.5]​?

To calculate the velocity travelled by the ball, differentiate the function.

dh/dt = 18 – 9.8t

Substitute t for 1 in the above Differential function

dh/dt = 18 – 9.8 (1)

But dh/dt = velocity

V = 18 – 9.8

V = 8.2 m/s

Average velocity = ( U + V ) / 2

Average velocity = (18 + 8.2)/2

Average velocity = 26.2/2

Average velocity = 13.1 m/s