Question

A ball is thrown up with the speed of 9.8 metre per second how high will it go before it fall ?​

Answer 1

Answer:

1.27 m

Explanation:

Given:

Initial speed (u) = 5 m/s

Final speed (v) = 0 m/s

Acceleration due to gravity (g) = -9.8 m/s²

To Find:

Height (h) reached by ball before it begins to fall

Explanation:

\begin{gathered} \sf From \ 3^{rd} \ equation \ of \ motion: \\ \boxed{ \bold{ {v}^{2} = {u}^{2} + 2gh}}\end{gathered}From 3rd equation of motion:v2=u2+2gh

Substituting values of v, u & g in the equation:

\begin{gathered} \sf \implies {0}^{2} = {5}^{2} + 2( - 9.8)h \\ \\ \sf \implies 0 = 25 - 19.6h \\ \\ \sf \implies 19.6h = 25 \\ \\ \sf \implies h = \frac{25}{19.6} \\ \\ \sf \implies h = 1.27 \: m\end{gathered}⟹02=52+2(−9.8)h⟹0=25−19.6h⟹19.6h=25⟹h=19.625⟹h=1.27m

Answer 2

Answer:

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