Question

two balls are projected simultaneously with the same speed from the top of a tower- one upward and the other downward. if there reached the ground in 6s and 2s, the height of the Tower is

Answer 1

Given:

Initial speed of projection = u

First ball reaches ground in t' = 6 s

Second ball reaches ground in t = 2s

To Find:

The height of the tower, h = ?

Formula Used:

Second equation of motion

$s=ut+0.5at^2$

Calculations:

For first ball:

$-h =ut'-0.5gt'^2$

For second ball:

$-h=-ut-0.5gt^2\\h=ut+0.5gt^2$

Add the above two equations to find the value of u:

$0=u(t'+t)+0.5g (t^2-t'^2)\\u = 0.5 g \frac{t'^2-t^2}{t'+t}\\u =0.5 g (t'-t)\\u = 0.5\times 9.8 \times (6-2) \\u = 19.6 m/s$

Substitute the value in any equation to find the value of h:

$h = 19.6\times 2+0.5 \times 9.8 \times (2)^2\\h = 39.2 +19.6 \\h =58.8 m$

Thus, the height of the tower is 58.8 m.

Learn more about: Projectile motion

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