If a ball is thrown vertically upward with a velocity of 128ft/s, then its height after t seconds is s=128t-16t^2.
a) what is the velocity of the ball when it is 240 ft above the ground on its way up? (consider pos direction)
b) what is the velocity of the ball when it is 240ft above the ground on its way down?
- I calculated maximum height reached by the ball which is 256 feet.
+118654 If a ball is thrown vertically upward with a velocity of 128ft/s, then its height after t seconds is s=128t-16t^2.
a) what is the velocity of the ball when it is 240 ft above the ground on its way up? (consider pos direction)
b) what is the velocity of the ball when it is 240ft above the ground on its way down?
- I calculated maximum height reached by the ball which is 256 feet. :) You didn't need to do this.
To start with it is symmetrical so whatever the velocity is going up it will be the same only minus going down.
To avoid confusion I am going to replace the s which is a height with the letter y.
\(y=128t - 16t^2\\ \dot y=128-32t\\ \text {you are being asked to find } \dot y \text{ when y=240 }\\ \text{But to do this you need to find t when y=240 first}\\ 240=128t-16t^2\\ 16t^2-128t+240=0\\ t^2-8t+15=0\\ (t-3)(t-5)=0\\ t=3\;\;or\;\;t=5\\ when\;\;t=3\\ \dot y=128-32*3\\ \dot y=128-96\\ \dot y=32\;feet/sec\\ \text{The velocity on the way up is } 32f/sec\\ \text{The velocity on the way down is } -32f/sec\\ \)
