A body is thrown with initial velocity 10 m/sec. at an angle `37^(@)` fro horizontal. Find
(i) time of flight
(ii) Maximum geight.
(iii) Range (iv) Position vector after `t=1` sec.

Answer

(i) 1.2 sec, (ii) 1.8 m, (iii) 9.6 m, (iv) `(16hat(i)-8hat(j))-(8hat(i)+hat(j))`

Explanation

(i) time of flight `T=(2u sin theta)/(g)=(2xx10xx3/5)/10=6/5=1.2` sec
(ii) Maximum height `H=(u^(2) sin^(2) theta)/(2g)=(100((9),(25)))/(2xx10)=9/5=1.8 m`
(iii) `R=(u^(2) sin 2 theta)/g=(100xx2xx3/5xx4/5)/10=240/25=9.6 m`
(iv) `x=8xx1=8 m`
`y=6xx1-1/2xx10xx(1)^(2)=1 m`
`vec(r)=8hat(i)+hat(j)`
Caution : This equation does ot given the horizontal distance travelled by a projectile when the final height is not the launch height.
Maximum Range
`R=(u^(2) sin 2 theta)/(g)`
for `theta=45^(@)`, R is maximum
`R_(max)=u^(2)/g`