A ball is thrown in the air from the top of a building. Its height, in meters above ground, as a function of time, in seconds, i s given by h(t)=−4.9x2+19.6x+24.
1 answer:
Answer:
a.2s b.43.6m
Step-by-step explanation:
h(t)=−4.9t2+19.6t+24
24m = inittial possition
-4.9 t2= -1/2 g t2
19.6t= Vo t
as the initial velocity is positive, the ball is thrown up.
the maximum height of the ball is reached when the velocity is 0
V= Vo-gt=19.6m/s-9.8m/s2 t=0
t=19.6/9.8=2s
<u>it takes 2seconds for the ball to reach its maximum height</u>
h(t)=−4.9t2+19.6t+24
h(2)=-19.6m+39.2m+24=43.6m
<u>the maximum height of the ball is 43.6m</u>
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