Ball is thrown verticaly upward from the roof 120ft.high(distance from roof at any time is s(t)=-16t^2 =43t)?

Question

After how many seconds will the ball hit the roof? if the ball on the way down were to fall over the side of the building, how long would it take for it to hit the ground?

Answer 1

i presume the equation you give should read s(t) = -16t^2 + 43t?

so by analogy to the general equation of s()=ut + 1/2 a t^2 implies that the initial velocity u is 43 m/s and the acceleration due to gravity is 1/2a = 16 so a =32.

the time to reach the top of its trajectory is when the final velocity is 0 so use v= u + at

implies 0=43+32t so t = 1.34375 to top. total time before hitting roof is twice this so 2.6875 s.

second part. . .  total height reached by ball is 120 + ( 43(1.34375) - 16(1.34375)^2) = 120 + 28.89 = 148.89 m

falling from this height from rest solve s = 0 + 1/2 a t^2

148.89 = 1/2 (32) t^2

t = 3.05 seconds to hit the ground