A body is projected at an angle of `30^(@)` with the horizontal and with a speed of `30ms^(-1)`. What is the angle with the horizontal after 1.5 second? `(g=10ms^(-2))`

Options

A. `0^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

A

Explanation

The time of light is given by
`T=(2u sin theta)/g=(2xx30xx1)/(10xx2)=3 sec`
Thus, after 1.5 sec the body is at the highest point. As the direction ofmotion is horizontal after 5 sec the angle with the horizontal is `0^(@)`.