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Sagot :
Answer:
a. 144m
b. After 2 secs
c. Ball hits the ground after 5 secs.
d. 128m
Step-by-step explanation:
A To find max height, find derivative of equation, let it equal zero, and sub in value into original equation: h(t) = -16t² + 64t + 80 h'(t)=-32t+64 -32t+64=0 32t=64 t=2 h(2) = -16(2²) + 64(2) + 80 h(2) = -64+128+80 = 144m B When h=144, t=2. Therefore, the ball reaches its max height after 2 secs. C When the ball hits the ground, h=0: -16t² + 64t + 80=0 t²-4t-5=0 (simplified) (t-5)(t+1)=0 t=5 and t=-1 Discount t=-1 (time cant be neg). Ball hits the ground after 5 secs. D Find h when t=1: h(t) = -16t² + 64t + 80 h(1) = -16(1²) + 64(1) + 80 h(1) = -16+64+80 h(1) = 128m.
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