Question

A ball is projected upwards from the foot of a tower.The ball crosses the top of the tower twice at an interval of 6s. the ball reaches the ground after 12s from the start. Find the height of the tower, take g=10m/s^2.

Answer 1

In a projectile the time taken to reach the highest point from the point of projection is equal to the time taken to fall from the highest point to the point of projection. That means the time of ascent = time of descent

If the total time taken is 12sec then time taken to reach the highest point is 6sec.

Similarly in 6sec it crosses the same point, let's say point A(top of tower), so it takes 3 sec to reach the highest point and 3more secs to come back to point A.
Then see the attachment for the calculation.


HOPE IT HELPS.... :-)

Answer 2

Answer:

135m

Explanation:

According to Newton's $1^{st}$ law of motion, v = u +at,

But here since the ball is projecting upwards

So, here a = -g (opposite of forced applied i.e Gravitational force)

Therefore, v = u -gt

Here v = 0, (Since the ball is projecting upward and returning to the ground hence, at the topmost point the velocity becomes zero)

By putting the values in above formula

we get,

0 = u - 10×6

=> u = 60 m/s

Now,

we know, s = ut + (1/2)a$t^{2}$

here a = -g (ball is projecting upward)

hence we get

s = ut - (1/2)g$t^{2}$

by putting the values

=> s = 60×3 - $\frac{1}{2}$×10×3×3

=> s = 180 - 45

=> s = 135 m

Hence, The height of the tower is 135 m.

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