From the information: v is 64 while c is 5
Differentiate the new equation h=-16[tex] t^{2} [/tex] + 64t + 5 to get [tex] \frac{dh}{dt} [/tex]= -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[tex] t^{2} [/tex] + 64t + 5. This gives
h=-16( [tex] 2^{2} [/tex]) + 64(2) + 5; h=-64+128+5=69. Hence h is 69ft
