Question

8 + 40.8t -{16t}^{2} " alt=" h(t) = 8 + 40.8t -{16t}^{2} " align="absmiddle" class="latex-formula">
at what height is the ball after 2 seconds?​

Answer 1

Answer:

$h(t) = 8 + 40.8t - {16t}^{2} \\ h(2) = 8 + 40.8(2) - 16 {(2)}^{2} \\ = 8+81.6 - 64 \\ = 89.6 - 64 = \boxed{25.6}$

After 2 seconds the ball will be at the height of 25.6 unit.

Answer 2

Step-by-step explanation:

When t = 2,

h(2) = 8 + 40.8(2) - 16(2)² = 8 + 81.6 - 64 = 25.6

Therefore the height after 2 sec is 25.6.