Question

A stone thrown vertically upward with an initial velocity of 40 ms⁻¹. Taking (g = 10 ms⁻²) draw the velocity-time graph of the motion of stone till it reaches back on the ground. Use graph to find the maximum height reached by the stone. What is the net displacement and total distance covered by the stone?

Answer 1

AnswEr:

Given,

• Initial velocity, u = 40 m $\sf{s}^{-1}$
• g = -10 m $\sf{s}^{-2}$ (upward motion )
• Final Velocity, v = 0

$\tt \: h = \frac{ {v}^{2} - {u}^{2} }{2g} \\ \\ \rightarrow \tt \: n = \frac{0 - {(40)}^{2} }{2( - 10)} \\ \\ \rightarrow \tt \: \frac{160 \cancel 0}{ - 2 \cancel 0} \\ \\ \rightarrow \: \tt h = 80 \: m$

Therefore, the maximum height reached by the stone will be 80 m.

• Net displacement on returning back = 0

Total distance covered -

$\tt \: 80 + 80 \\ \\ \tt= 160 \: m$

Answer 2

Explanation:

Given,

Initial velocity, u = 40 m \sf{s}^{-1}s

−1

g = -10 m \sf{s}^{-2}s

−2

(upward motion )

Final Velocity, v = 0

[TEX]\begin{lgathered}\tt \: h = \frac{ {v}^{2} - {u}^{2} }{2g} \\ \\ \rightarrow \tt \: n = \frac{0 - {(40)}^{2} }{2( - 10)} \\ \\ \rightarrow \tt \: \frac{160 \cancel 0}{ - 2 \cancel 0} \\ \\ \rightarrow \: \tt h = 80 \: m\end{lgathered}[/TEX]

Therefore, the maximum height reached by the stone will be 80 m.

• Net displacement on returning back = 0

Total distance covered -

$\begin{lgathered}\tt \: 80 + 80 \\ \\ \tt= 160 \: m\end{lgathered}$

=160m