Answer:
Step-by-step explanation:
a)
When the height reaches the maximum, the 1st differentiation of h(t) [h'(t)] = 0
[tex]h(t)=-16t^2+52t+48[/tex]
[tex]h'(t)=0[/tex]
[tex]2(-16t^{(2-1)})+52t^{(1-1)}+48(0)=0[/tex]
[tex]-32t+52=0[/tex]
[tex]t=\frac{13}{8} \ s[/tex]
[tex]h(\frac{13}{8} )=-16(\frac{13}{8}) ^2+52(\frac{13}{8}) +48[/tex]
[tex]=\boxed{90\frac{1}{4} \ ft}[/tex]
b)
When the ball hits the ground, h(t) = 0
[tex]h(t)=0[/tex]
[tex]-16t^2+52t+48=0[/tex]
[tex]-4t^2+13t+12=0[/tex]
[tex](-4t-3)(t-4)=0[/tex]
[tex]t=-\frac{3}{4} \ or\ 4[/tex]
Since t is positive, therefore [tex]\boxed{t=4\ s}[/tex]
