Question

A ball is thrown vertically downwards from the top of tower with speed 20 m/s reach the ground in 4 s. The height of tower is (g = 10 m/s2)​

Answer 1

Answer

The initial velocity of the ball is =20 m/s.

The final velocity of the ball is =80 m/s.

The acceleration due to gravity is =10 m/s 2

So, the height of the tower can be obtained from the expression:

v 2 = u2 + 2gh

Substitute the values in above expression:

80 2   = 20 2  + (2×10×h)

h = $\frac{6400 − 400}{20}$

h = $\frac{600}{200}$ ⇒ 300 m

Answer 2

$\large\underline{\sf{Given}}$

» A ball is thrown vertically downwards from the top of tower.

» Velocity (u) of ball = 20 m/s.

» Time taken by ball to reach the ground = 4 s.

» Acceleration (g) of ball = 10 m/s².

$\large\underline{\sf{To\:Find}}$

» Height of the tower?

$\large\underline{\sf{Solution}}$

$\underbrace{\underline{\sf{\bigstar\:Understanding\:the\:concept\:::}}}$

Here, we are given that a ball is thrown vertically downwards from a tower. We have velocity of ball, time taken by to reach the ground and acceleration of ball. We have to find out the height of the tower. So, we will use formula :: h = ut + ½ gt². In which g means that the object is in freefall under the effect of gravity, which means that acceleration of object is 'g', 'u' is speed of the object and 't' is the time taken to travel the distance 'h'. So, let's solve it!

$\sf \longrightarrow\: h = ut + \dfrac{1}{2}$

Putting all known values,

$\sf \longrightarrow\: h = \big(20\:\times\:4\big) + \dfrac{1}{2}\:\times\:10\:\times\:\big(4\big)^2$

$\begin{gathered}\\ \sf \longrightarrow\: h = 80 + \dfrac{1}{2}\:\times\:10\:\times\:4\:\times\:4\end{gathered}$

$\begin{gathered}\\ \sf \longrightarrow\: h = 80 + \dfrac{1}{\cancel{2}}\:\times\:10\:\times\:\cancel{16}\end{gathered}$

$\begin{gathered}\\ \sf \longrightarrow\: h = 80 + 10\:\times\:8\end{gathered}$

$\begin{gathered}\\ \sf \longrightarrow\: h = 80 + 80\end{gathered}$

$\begin{gathered}\\ \sf \longrightarrow\:\pink{ Height\:of\:tower\:(h) = 160\:m}\end{gathered}$

$\therefore\:{\underline{\textsf{\textbf{Option\:(c)\:160\:m\:\sf{is\:correct!}}}}}$

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