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ball $1$ has traveled for $(t+2)s$

Both balls traveled same dist.

$∴∣dfrac12(t+2)_{2}=ut+21 g_{(t)_{2}}$

$21 ×10(t+2)_{2}=40t+2t ×10(t)_{2}$

$5(t+2)_{2}=40t+56_{2}$

$5(t_{2}+4+4t)=40t+5t_{2}$

$h=21 gt_{2}$ for $1$st ball

$h=40(t−2)+21 g(t−2)_{2}$ $2$nd ball

$21 gt_{2}=40(t−2)+21 (t−2)_{2}$

$40(t−2)=21 g(t_{2}−(t−2))_{2}$

$4t−8=2t−2⇒t=3$

$h=21 gt_{2}$

$=21 ×10×(3)_{2}=45m$

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