A projectile has a range of ` 60 m` and reaches a maximuum height of `12 m`. Calculate the angle at which the projectile is fired and initial velcity of projection of projectile. Given `g=10 ms^(-2)`.

Explanation

Here, `R=60 m, H=12 m, theta =?`
Horizontal range,
`R=(2u^(2) sin theta cos theta)/g =60` …9i)
Maximum hight , `H =(u^(2) sing^(2) theta)/ 2 g) =12` ..(ii)
:. `H/R =1/4 tan theta =(12)/(60) -1/5 =0.2`
or tan `theta 0.8 =tan 38^(@) 40' or theta 38^(@) 40'`
From (ii), `u^(2) sin^(2) theta =12 xx 2 xx g`
`12 xx 2 xx 10 =240`
`u sin theta sqrt 240 =15.49`
`u =(15 .49)/(sin theta) =(15 .40)/(sin 38^(@) 40') = (15 .49)/(0.6248) = 24.8 ms^(-1)`.