Asked by sandy
If a ball is thrown vertically upward from the roof of 48 foot building with a velocity of 80 ft/sec, its height after t seconds is s(t)=48+80t–16t2. What is the maximum height the ball reaches? (my answer is 148). What is the velocity of the ball when it hits the ground (height 0)?
Answers
Answered by
Reiny
your first answer is correct.
for the second part, first find when the ball hits the ground, that is, set s(t) = 0
16t^2 - 80t - 48 = 0
t^2 - 5t - 3 = 0
t = (5 ± √37)/2 = 5.541 or a negative
sub 5.541 into your velocity, which you obviously had correct.
for the second part, first find when the ball hits the ground, that is, set s(t) = 0
16t^2 - 80t - 48 = 0
t^2 - 5t - 3 = 0
t = (5 ± √37)/2 = 5.541 or a negative
sub 5.541 into your velocity, which you obviously had correct.
Answered by
sandy
i did however my answer is still wrong. can you show me step by step. this way i can see what i am doing wrong. thank you
Answered by
Rool
the answer's -97.32
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