A particle is projected into space at an angle of `60^(@)` with an initial speed of `400ms^(-1)` .Calculate the instantaneous velocity of the projectile along the horizontal, vertical and the resultant velocity at t=15s. Calculate instantaneous position of the particle along the horizontal and vertical `(g=9.8ms^(-2))`

Explanation

Given, `v_0=400ms^(-1),theta=60^(@),t=15s`
We know that `v_y=(v_0sintheta)-"gt"`
i.e. `v_y=400xxsin60-9.8xx15=(400xx0.866)-147=346.4-147`
`v_y=199.4ms^(-1)" "v_(0x)=v_x=v_0costheta=400xxcos60`
`:." "v_x=400xx0.5=200ms^(-1)`
Resultant velocity `v=sqrt(v_x^2+v_y^2)`
i.e `v=sqrt((199.4)^2+(200)^2)" "v=282.4ms^(-1)`
Horizontal distance covered `x=(v_0costheta)t=200xx15`
x=3000m
vertical distance covered `y=(v_0sintheta)t-1/2g t^2=(346.4)15-1/2xx9.8xx225`
`=5196-1102.5" "y=4093.5 m`
Position co-ordinate=(x,y)=(3km,4.1km)