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?

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

To find the height of the tower, we need to use the equation of motion for vertical motion with constant acceleration:

h = u*t + (1/2)*g*t^2

Where:
h = height of the tower
u = initial velocity of the ball (thrown vertically upwards) = 100 m/s
t = time taken for the ball to reach the pond = 25 seconds
g = acceleration due to gravity (approximately 9.8 m/s^2)

Let's substitute the given values into the equation and solve for h:

h = (100 m/s)*(25 s) + (1/2)*(9.8 m/s^2)*(25 s)^2
h = 2500 m + (1/2)*(9.8 m/s^2)*(625 s^2)
h = 2500 m + (1/2)*(9.8 m/s^2)*(625 s^2)
h = 2500 m + 3062.5 m
h = 5562.5 m

Therefore, the height of the tower is approximately 5562.5 meters.

None of the options provided in the question match this answer. It's possible that there is a mistake in the given information or in the options.

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.

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