1.

The horizontal range of a projectile fired at an angle of 15^(@) is 50 m. If it is fired with the same used at angle of 45^(@), its range will be

Answer»

125 m
75m
100 m
50 m

Solution :Horizontal RANGE `R= (u^2 sin 2THETA)/(g)`
If the equal speed
`(R_1g)/(sin 2theta_1) = (R_2g)/(sin 2 theta_2)`
`R_(2) = (2xx sin 90^@)/(sin 30^@) = 100 `m


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