1.

The maximum height attain by a projectile is increased by ` 10%` by increasing its speed of projection, without changing the angle of projection. What will the percentage increase in the horizontal range.

Answer» As max. height ,
`H=u^(2)/ 2g) sin ^(2) theta` …(i)`
Let `Delta H` be the increase in (H) when (u) changes by `Delta u` . Differentiating (i), we get
`DeltaH` =(2 u Delta u sin^(2) theta)/(2g)`
:. (DeltaH)/H =(2 Delta u)/u`
Given, % increase in (H) is `10%,
so `(Delta H) /H (10)/(1000) =0.1. Therfore, (2 Deltau)/u =0.1
As `R= (u^(2) sin2 theta )/g`
Therfore, Delta R =(2 u Delta u)/g xx sin 2 theta`
And `(Delta R)/R (2 Delta u)/u =0.1`
:. `% increase in horizontal tange
`(Delta R)/R xx 100 =0.1 xx 100 =10%`.


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