Question

a stone is projected vertically up from the top of a tower at a velocity of 9.8ms-1 . It reaches the foot of the building in 4second .what is the height of the tower and the velocity of the stone when it reaches the ground ? g= 9.8ms-2

Answer 1

For calculation of Height use

H= u*t +1/2g*t^2

And then for velocity

V^2=u^2+2gH

Answer 2

Answer:

part (a) Height of the building = h = 94.1 m

Part (b) v = 20.62 m/s.

Explanation:

Given,

• Time of flight = T = 4 sec
• g = $9.81\ m/s^2.$

Part (a)

Let u be the initial speed of the stone. Assuming downward direction of the motion is positive.

Therefore acceleration of the stone = a = -g

$\therefore T\ =\ \dfrac{2u}{g}\\\Rightarrow u\ =\ \dfrac{Tg}{2}\\\Rightarrow u\ =\ \dfrac{4\times 9.81}{2}\\\Rightarrow u\ =\ 18.62\ m/s.$

Now let us assume that h be the height of the tower. Hence the total displacement of the stone in the given time of flight is equal to the height of the tower in a downward direction.

$\therefore h\ =\ ut\ +\ \dfrac{1}{2}gt^2\\\Rightarrow h\ =\ 18.62\times 4\ +\ 0.5\times 9.81\times 4^2\\\Rightarrow h\ =\ 94.1\ m$

Part (b)

Let v be the final speed of the stone when it reaches the ground.

From the equation of kinematics,

$v^2\ =\ u^2\ +2gh\\\Rightarrow v\ =\ \sqrt{u^2\ +\ 2gh}\\\Rightarrow v\ =\ \sqrt{18.62^2\ +\ 2\times 9.81\times 94.1}\\\Rightarrow v\ =\ 46.82\ m/s$.