Question

stone is dropped from the top of the tower and reaches the ground in 3s. Then
the height of the tower is (g = 9.8 m/s)
A) 18.6m
(B) 39 2m (C) 44.1m (D) 98m​

Answer 1

Answer:

44.1 m.

C. option is correct.

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Acceleration ( g ) = 9.8 $m/sec^2$

Time ( t ) = 3 sec

First find the final velocity

we have first equation of motion

v = u + g t

v = 0 + 9.8 × 3 m / sec

v = 29.4 m / sec

Now we have third equation of motion

$\displaystyle v^2=u^2+2gh$

Put the value here  we get

$\displaystyle (29.4)^2=0^2+2\times9.8\times h\\\\\displaystyle (29.4)^2=19.6h\\\\\displaystyle h =\frac{864.36}{19.6} \ m\\\\\displaystyle h =44.1 \ m$

Thus the height of tower is 44.1 m.

Answer 2

Answer:

Height of the tower from which the stone was thrown = 44.1 m

Step by step explanations :

gívєn thαt,

stone is dropped from the top of the tower and reaches the ground in 3s.

Here,

We have,

initial velocity of the stone = 0 m/s [because ball was dropped]

time taken to reach the ground = 3 seconds

let the height of the tower be h

now,

we have,

initial velocity(u) = 0 m/s

time taken(t) = 3 s

height = h

gravitational acceleration(g) = -9.8 m/s²

so,

by the gravitational equation of motion,

S = ut + ½gt²

putting the values,

S = 0(3) + ½(-9.8)(3)(3)

S = -44.1 m

negation of height shows the height downwards

so,

Height of the tower from which the stone was thrown

= 44.1 m

__________________

Correct option :

(C) 44.1 meter