If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t2.
What is the velocity of the ball when it is 96 ft above the ground on its way up? (Consider up to be the positive direction. Round the answer to one decimal place.)
What is the velocity of the ball when it is 96 ft above the ground on its way down? (Round the answer to one decimal place.)
*Revised*
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2.
What is the velocity of the ball when it is 96 ft above the ground on its way up? (Consider up to be the positive direction. Round the answer to one decimal place.)
What is the velocity of the ball when it is 96 ft above the ground on its way down? (Round the answer to one decimal place.)
velocity= ds/dt= 160-32t
use the s formula to find t (you get two t, a quadratic) when s=96
then put those times in the velocity equation above.
that is what i am having trouble with =/
i got 96= 160t-16t^2 and i don't know how to factor. so i tried -16t^2+160t-96 and used the quadratic formula and i got two answer: .6 and 9.4...i do not know what i am doing wrong, but i really need help finding t when it's 96 feet.
Thank you.
To find the velocity of the ball when it is 96 ft above the ground on its way up and down, we need to find the derivative of the height function with respect to time (t).
1. On its way up:
To find the velocity of the ball when it is 96 ft above the ground on its way up, we need to find the time (t) at which the height (s) is 96 ft.
Given the height equation s = 160t - 16t^2, we can set s equal to 96 and solve for t:
96 = 160t - 16t^2
Rewriting the equation in the standard quadratic form:
16t^2 - 160t + 96 = 0
Next, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Once we find the value of t, we can substitute it back into the equation to find the velocity.
2. On its way down:
To find the velocity of the ball when it is 96 ft above the ground on its way down, we need to find the time (t) when the ball is at the maximum height and then find the time at which the height is 96 ft.
The height function s = 160t - 16t^2 is a quadratic equation in terms of time. The maximum height occurs at the vertex of the parabolic curve, and the formula for finding the vertex of a quadratic equation is -b/2a.
In this case, a = -16 and b = 160, so the time at the maximum height can be found using -b/2a:
t_vertex = -b / 2a = -160 / 2(-16) = 5 seconds
Once we have the time at the maximum height, we can determine the velocity when the ball is 96 ft above the ground by finding the velocity at t = 5 seconds using the derivative of the height function.
d/dt (160t - 16t^2) = 160 - 32t
Substituting t = 5 into the derivative equation:
v = 160 - 32(5)
v = 160 - 160
v = 0 feet per second
The velocity of the ball when it is 96 ft above the ground on its way down is 0 ft/s.