Question

A body is projected with a velocity of 40ms^-1.After 2s,it crosses a vertical pole of length 20 calculate the angel of projection and horizontal range, g=10 ms^-2

Answer 1

$u=40m/s$ g=10m/s2g = 10 m/s^2g=10m/s2

At t1=2sect_1 = 2 sect1​=2sec $y=20.4m$
$y=usin\theta t-gt²/2$
$20.4=40sin\theta \times 2-g2²/2$
=> $sin\theta =1/2$
(1)=> θ=30° \theta = 30°angle of projection, θ=30°

$\left(2\right)\Rightarrow$Horizontal range = R = u² Sin2Ф / g = 40² sin60° / 10 $\cong 138m$