Answer:
For the ball thrown vertically
upwards, the time taken by the ball to come back is, \[{{T}_{1}}=2{{\mu
}_{1}}/g\].
For the ball
projected at an angle 9 with their vertical, the time of flight is, \[{{T}_{2}}=2{{\mu
}_{2}}\cos \theta /g\]
Since time for both
the balls is same, so \[\frac{2{{u}_{1}}}{g}=\frac{2{{u}_{2}}\cos \theta }{g}\]or
\[{{u}_{1}}={{u}_{2}}\cos \theta \].
Now, \[{{h}_{1}}=\frac{u_{1}^{2}}{2g}\]and
\[{{h}_{2}}=\frac{u_{2}^{2}}{2g}{{\cos }^{2}}\theta \]; Hence, \[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{u_{1}^{2}}{u_{2}^{2}{{\cos
}^{2}}\theta }=\frac{u_{2}^{2}{{\cos }^{2}}\theta }{u_{2}^{2}{{\cos }^{2}}\theta
}=1\]
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