A ball is thrown vertically upward with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s = 48 t - 16 t ^2. (a) At what time t will the ball strike the ground? (b) For what time t is the ball more than 32 feet above the ground?

Question

Garrett

9 February, 13:56

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Answers (1)

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Romeo Hicks · Answer 1

Hello,

s=48t-16t²

a)

s=0==>16t (3-t) = 0==>t=0 or t=3

b)

s>32==>48t-16t²>32===>16 (t²-3t+2) <0

Δ=9-8=1

==>16 (t-1) (t-2) <0

==>1