Question

A potato is launched verticaily upward with an initial velocity of 105ft/s from a potato gun at the top of a 90ft tall building. The helght of the potato (in feet) above the ground after t seconds is given by the function s(t)=−16t 2+105t+90. The velocity of the potato at t=0.5 seconds is [t/s and at t=5.5 seconds is it/s The speed of the potato at t=0.5 seconds is t/s and at t=5.5 seconds is ft/s While airborne, the height of the potato is increasing on the open interval and decreasing on open the interval

Answer 1

To find the velocity of the potato at a given time, differentiate the equation for height with respect to time. At t = 0.5 seconds, the velocity is 89 ft/s and the speed is also 89 ft/s. At t = 5.5 seconds, the velocity is -5 ft/s and the speed is 5 ft/s. The height of the potato is increasing from 0.5 seconds to 5.5 seconds.

Step-by-step explanation:

To find the velocity of the potato at a given time, we can differentiate the equation for height with respect to time. The derivative of s(t) = -16t^2 + 105t + 90 is v(t) = -32t + 105. Substituting t = 0.5 seconds, we get v(0.5) = -32(0.5) + 105 = 89 ft/s. Substituting t = 5.5 seconds, we get v(5.5) = -32(5.5) + 105 = -5 ft/s.

The speed of the potato is the absolute value of the velocity. So, at t = 0.5 seconds, the speed is |v(0.5)| = |-32(0.5) + 105| = 89 ft/s. At t = 5.5 seconds, the speed is |v(5.5)| = |-32(5.5) + 105| = 5 ft/s.

The height of the potato is increasing when the velocity is positive, and decreasing when the velocity is negative. Since at t = 0.5 seconds the velocity is positive (89 ft/s) and at t = 5.5 seconds the velocity is negative (-5 ft/s), the height of the potato is increasing from 0.5 seconds to 5.5 seconds.