a-projectile-is-launched-vertically-upwards-and-its-height-above-the-ground-afte

Question

A projectile is launched vertically upwards and its height above the ground after ﻿ttt﻿ seconds is given by the function ﻿h(t)=50t−5t2 h(t) = 50t - $$5t^2 h(t)=50t$$−5t2﻿, where ﻿0≤t100\le t0≤t10﻿. At what time is the projectile at its maximum height?

Accepted Answer

t=5t=5 ss

Answer

t=10t=10﻿ ss

Answer

t=15t=15﻿ ﻿sss﻿

Answer

﻿t=20t=20 ss

A projectile is launched vertically upwards and its height above the ground after ttt seconds is given by the function h(t)=50t−5t2 h(t) = 50t - 5t^2 h(t)=50t−5t2, where 0≤t<100\(\le\) t<100≤t<10. At what time is the projectile at its maximum height?

A projectile is launched vertically upwards and its height above the ground after $t$ seconds is given by the function $h(t) = 50t - 5t^2$ , where $0\(\le\) t<10$ . At what time is the projectile at its maximum height?