The speed with which a bullent can be fired is `150 ms^(-1)`. Calculate the greatest distance to which it can be projected and also the maximum height to which it would rise .

Explanation

The greatest horizontal range is achieved at an angle of projection of `45^(@)`
Components of initial velcoity `= 150 cos 45^(@)C = 106.06 ms^(-1)` .
Now , if T is the time of flight , then considering the vertical motion of the bullet ,
` mu = 106.06 ms^(-1), a = - 9.8 ms^(-1) ,s = 0 , t =T`
Using `s = ut+ (1)/(2) at^(2)`
we get `0 = 106.06 T - (1)/(2) 9.8 T^(2)rArr T = (106.06 xx 2)/(9.8) = 21.64 sec`,
`therefore ` Maximum Horizontal range = horizontal component of velocity `xx` total time of flight
` =106.06 xx 21.46 = 2295. 14 m`
Agin , if `H_("max")` be the maximum height of which the bullet rises then
` u = 106.06 ms^(-1), a = - 9.6 ms^(-1)`
`v = 0,s = H_(max) = (u^(2))/(2g) rArr H_(max) = ((106.06)^(2))/(19.6) = 573.91m`