Question

A ball is freely dropped from a tower,and it reaches the ground in 4sec ...calculate the distance of tower ??? ​

Answer 1

$given \: time \: is = 4secods \\ we \: have \: to \: find \: the \: length \: of \: tower \\ then \: by \: second \: equation \\ s = ut - (1÷2)gt {}^{2} \\ s = 0 \times t - (1÷2)( - 9.8) \times 4 \times 4 \\ s = 78.8m$

Answer 2

Height of the tower is 78.8 metres.

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Let the length of Tower be S metres.

Now,

The ball in freely falling in the direction of the Gravitational Force.

Thus,

Acceleration due to gravity(g) will be acting on the ball.

So,

$\boxed{\sf{\red{g = 9.8 m/sec^{2}}}}$

$\rule{200}{2}$

Initial velocity (u) = 0 m/sec

Time taken to cover distance (t) = 4 sec

Acceleration (g) = 9.8 m/sec²

$\rule{200}{2}$

We know that :-

$\boxed{\bold{\red{\underline{s = ut + \frac{1}{2}gt^{2}}}}}$

$\rule{200}{2}$

Height of the tower will be as follows :-

$s = 0 \times 4 + \frac{1}{2} \times 9.8 \times 16 \\ \\ \\ s = 9.8 \times 8 \\ \\ \\ s = 78.8m$

$\rule{200}{2}$

Height of tower is 78.8 metres

$\rule{200}{2}$

NOTE:- If we will take "g" as 10 m/sec², The height of tower will come 80 metres.

$\rule{200}{2}$