Question

a ball is projected vertically upward from foot of tower.the ball crosses the top of tower after interval of 6 sec. and reaches ground after 12 sec. what is the height of the tower?
please answer quick because the question can give you 15 points
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Answer 1

Answer:

Time of flight for 1st journey is

s = ut + 1/2at^2

Displacement will be zero

s = u × 6 + 1/2 × 10 × 6 × 6

here s isn't distance it's displacement since the ball will come again at the same point on the top of the tower after 6 sec so from here we will calculate u

0 = 6u + 1/2 ×(- g ) × 6 ×6

0 = 6u × (-180)

6u = 180

u = 30m/s

Now we can go for 2nd journey

We have to calculate height of tower so we will apply 3rd equation of motion

V^2 - U^2 = 2as

Final point will be ground i.e final velocity = 0

- 30 × 30 = 2 × 10 × (-s)

s is taken with negative sign showing we are moving downwards

30×30 = s

2×10

S = 15 × 3 = 45 m

so the height of the tower is 45m

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