Question

a ball is thrown up with a speed of 15m/s how high will it go before it begins to fall​

Answer 1

$\huge\underline\mathrm\blue{✧Solution:-}$

☃️ $\mathrm\green{Given➝}$

$\pink\rightarrow{ }$$\mathrm\orange{Initial\:velocity\:=\:15\:m/s}$

$\pink\rightarrow{ }$$\mathrm\orange{Acceleration\:due\:to\: gravity(g)=}$$\mathrm\orange{-10}$$\mathrm\orange{\:\:\:\:\:\:\:\:ms^{2}}$ $\mathrm\green{(since\: the\: ball\: is \:going \:against}$$\mathrm\orange{\:\:\:\:\:\:\:}$$\mathrm\green{the\:gravity.)}$

$\pink\rightarrow{ }$$\mathrm\orange{Final\:velocity\:=\:0\:m/s}$

$\pink\rightarrow{ }$$\mathrm\orange{Height(s)=\:To\:find}$

☃️ $\mathrm\green{We\: know\:,}$$\mathrm\orange{v^{2}}$$\mathrm\orange{=}$$\mathrm\orange{\:u^{2}}$$\mathrm\orange{+}$$\mathrm\orange{\:2as}$

☃️ $\mathrm\green{putting\:the\:given\: values\:in\:equation,}$

$\pink\rightarrow{ }$$\mathrm\orange{0^{2}}$$\mathrm\orange{=}$$\mathrm\orange{\:(15)^{2}}$$\mathrm\orange{+}$$\mathrm\orange{\:2(-10)(s)}$

$\pink\rightarrow{ }$$\mathrm\orange{0\:=\:225\:+\:(-20)(s)}$

$\pink\rightarrow{ }$$\mathrm\orange{0\:=\:225\:-20s}$

$\pink\rightarrow{ }$$\mathrm\orange{20s\:=\:225}$

$\pink\rightarrow{ }$$\mathrm\orange{s}$$\mathrm\orange{=}$$\orange{\frac{225}{20}}$

$\pink\rightarrow{ }$$\mathrm\orange{s}$$\mathrm\orange{=}$$\mathrm\orange{11.25\:m}$

$\normalsize\underline\mathrm\blue{Therefore\:the\:ball\:will\:go\:upto\:11.25\: metres,}$$\normalsize\underline\mathrm\blue{before\:it\:starts\: falling.}$