A ball is thrown upwards with a velcoity of ` 80 m//s` at an angle of ` 30^@` to the horizontal . Find its velcituy after one second.

Explanation

Let `v_x, v_y` be the rectanular component velocities fo ball at a point agter ` 1` second of projection, where` v_x= ucos 30^@`
` = 80 xx sqrt 3//2 = 40 sqrt 3 ms^(-1)`
Taking vertical upward motion of ball for ` 1sec` we have
` v_y = ? U_y =u sin 30^2= 80 xx 1//2 = 40 m//s`,
a_y =- 9.8 ma^(-2) , t= 1 s`
As, ` v_y =u_y +a_y t=40 + (- 9.8) 1`
`= 40 - 9.8 = 30.2 ms^(-1)`
` Resultant velocity ` v= sqrt ( v_x^2 +v_y^2)
` =sqrt ((40 sqrt3)^2 + (30.2)^@) = 75.57 ms^(-1)`
` If ` theta` si the angle which the resultant velocity makes with the x-axis then ltBRgt ` tan theta = v_y/v_x = ( 30.2)/( 40 sqrt3) = 0. 4359`
= tan 23^@ 33' ` or ` theta = 23^@ 33' `.