Question

How to do number 13 and 14

Answer 1

From the information: v is 64 while c is 5

Differentiate the new equation h=-16$t^{2}$ + 64t + 5 to get $\frac{dh}{dt}$= -32t + 64.

no 13). At maximum height this derivative equals zero so: -32t + 64 = 0;  -32t = -64; t=2.Hence ans is 2 secs

no 14). put t as 2 sec in the equation: h=-16$t^{2}$ + 64t + 5.  This gives

h=-16( $2^{2}$) + 64(2) + 5; h=-64+128+5=69. Hence h is 69ft