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(1) h(t)=-4t²+16t+20=0.

This can be written 4t²-16t-20=4(t²-4t-5)=0.

4(t-5)(t+1)=0, so t=5 seconds for the ball to hit the ground, because t-5=0, so t=5. (The other “solution” would be t+1=0, t=-1, which is one second before the ball was kicked! So we reject t=-1.)

(2) We have to solve h(t)=28, so:

-16t²+32t+48=28, 16t²-32t+28-48=

16t²-32t-20=0=4(4t²-8t-5)=4(2t+1)(2t-5). To make this zero we use 2t-5=0, so 2t=5, t=5/2=2.5 (seconds). Just type 2.5 in the box.

(3) h(t)=-16t²+16t+32=20,

16t²-16t+20-32=16t²-16t-12=0=4(4t²-4t-3)=4(2t-3)(2t+1).

To make this zero, 2t-3=0, 2t=3, t=1.5.

(4) h(t)=-16t²+64t-64=0, 16t²-64t+64=0=16(t²-4t+4)=16(t-2)².

So t-2=0, t=2 seconds for the rocket to fall to the ground.

ago by Top Rated User (804k points)

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