Question

5. A projectile is fired vertically upwards with height h given by: h = 25t - 4.9^2 where 0 ≤ t ≤ 5

Answer 1

if we need find:$h(5)=25*5-4.9*5*5=25(5-4.9)=2.5$

if we need find$h_{max},h'(t)=25-2*4.9*t$ decreases,$h'(t_*)=0$ ,

$t_*=\frac{25}{2*4.9}$ ,$\implies h_{max}=h(t_*)=25*\frac{25}{2*4.9}-4.9*\frac{25}{2*4.9}*\frac{25}{2*4.9}=$

$=\frac{625}{2*4.9}(1-\frac{1}{2})=\frac{625}{4*4.9}$

if we need find$t_{1,2}|h(t_{1,2})=0,h(t)=t*(25-4.9t)=0\implies t_1=0,t_2=\frac{25}{4.9}$