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)?

Answer 1

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.

Answer 2

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

Answer 3

the answer's -97.32