1.

Figure shows the trajectory of a projectile fired at an angle theta with the horizontal. The elevation angle of the highest point as seen from the point of launching is varphi. The relation between varphi and theta is :

Answer»

`tanvarphi=1/2tantheta`
`tan^(2)varphi=1/2tan^(2)THETA`
`sinvarphi=1/2sintheta`
`COS^(2)varphi=1/2cos^(2)theta`

Solution :Here `tanphi=H/(R/2)=((u^(2)SIN^(2)theta)/(2g))/((2U^(@)sinthetacostheta)/(2g))=(sintheta)/(2costheta)`
`tanphi=1/2tantheta`


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