a ball is thrown vertically upwards from the top of a tower with a speed of 100 ms-1. it strikes the pond near the base of the tower after 25 seconds. the height of the tower is?

Question

a)500m b)125m c)625m

Answer 1

Tr=-Vo/g = -100/-9.8=10.20 s.=Rise time.

d = Vo*Tr + 0.5g*Tr^2
d = 100*10.20 - 4.9*10.20^2 = 510 m Above tower.

Tf = 25 - 10.20 = 14.8 s. = Fall time.

h+d = 0.5g*Tf^2 = 4.9*14.8^2 = 1073 m.
h + d = 1073
h + 510 = 1073
h = 1073-510 = 563 m. = Ht. of the tower

Answer 2

Correction:

Tr = -Vo/g = -100/-10 = 10 s.=Rise time.

h1 = Vo*Tr + 0.5g*Tr^2
h1 = 100*10 - 5*10^2=500 m. Above tower.

Tf = 25 - 10 = 15 s. = Fall time.

h1+h2 = 0.5g*Tf^2 = 5*15^2 = 1125 m.
h1+h2 = 1125
500 + h2 = 1125
h2 = 1125-500 = 625 m.=Ht. of the tower.