Respuesta :
We need derivative of h(t)
[tex]\\ \sf\longmapsto \dfrac{d}{dx}-16t^2+82t+96[/tex]
[tex]\\ \sf\longmapsto -32t+82[/tex]
- Now t=0
[tex]\\ \sf\longmapsto h(0)[/tex]
[tex]\\ \sf\longmapsto -32(0)+82[/tex]
[tex]\\ \sf\longmapsto 82m[/tex]
Hence
.[tex]\\ \sf\longmapsto H_{max}=96+82=178m[/tex]
Answer:
[tex]{ \rm{h(t) = 96 + 80t - 16 {t}^{2} }} \\ \\ { \rm{ \frac{dh}{dt} = 0 + 80 - (2 \times 16)t}} \\ \\ { \rm{ \frac{dh}{dt} = - 32t + 80}}[/tex]
• At maximum height, dh/dt is 0
[tex]{ \rm{ - 32t + 80 = 0}} \\ \\ { \rm{ - 32t = - 80}} \\ \\ { \rm{t = 2.5 \: seconds}}[/tex]
• Maximum height = 96 + (80 × 2.5) - (16 × 2.5²)
Max. height = 196 meters
Answer:
[tex]{ \rm{h(t) = 96 + 80t - 16 {t}^{2} }} \\ \\ { \rm{ \frac{dh}{dt} = 0 + 80 - (2 \times 16)t}} \\ \\ { \rm{ \frac{dh}{dt} = - 32t + 80}}[/tex]
• At maximum height, dh/dt is 0
[tex]{ \rm{ - 32t + 80 = 0}} \\ \\ { \rm{ - 32t = - 80}} \\ \\ { \rm{t = 2.5 \: seconds}}[/tex]
