Answer:
As is known from
theory,
Total
KE of the ball \[=\frac{7}{10}m{{\upsilon }^{2}}=\frac{7}{10}\times 1\times
{{\left( 20 \right)}^{2}}=280J\]
As work done in stopping the ball = K.E. of ball
\[\therefore \] \[(mg\,\sin
\theta )\times s=280\]
\[s\,=\frac{280}{mg\,\sin
\,\theta }=\frac{280}{1\times 9.8\times \frac{1}{2}}=57.14\,m\]
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