A projectile takes off with an initial velocity of `10 m//s` at an angle of elevation of `45^@`. It is just able to clear two hurdles of height 2 m each, separated from each other by a distance d. Calculate d. At what distance from the point of projection is the first hurdle placed? Take `g = 10m//s^2`.

Explanation

Here `u= 10 m//s, theta = 45^@`. ltBrgt ` AA-1= BB-1 =2m`,
.
Let the projectile projected from (O), clears the first hurdle ` AA_1` in time (T),
Taking the vertical motion of rpojectile from (O) to (A), we have , ` y m, y-0 =0`,
` u-y = u sin 45^2 = 10 xx 1/( sqrt2) =5 sqrt 2 m//s ,`
` a_y =- 10 m//s^2, t+ 1/2 a_y t^2` ltBrgt As ` y=y -0 +u_y t+ 1/2 t^2`
:. ` 2= 0 + 5 sqrt 2 t + 1/2 ( - 10) t^2`
or `5 t^2 - 5 sqrt 2 t + 2=0`
On solving we ger, `t=0. 3908` or 1.0232` second, It means the projectile crosses the first hurdle after time ` 0.3908 s` and second hurdle after tiem ` 1. 0232` second.
:. Time taken by projectile to go from one hurdle to another (t_2-t-1) = 1. 0232- 0.3908`
= 0.6324 s`
Hence distance ` A_1 B_1,
`r=u cos 54^2 xx (t-2- t_1) ` ltBrgt `= 10 xx 1/(sqrt 2) xx 0.6324 = 4.472 m`
Distance `OA_1= u cos 45^2 xxt_1`
`= 10 xx 1/9sqrt 2) xx 0. 3908 = 2 .764 m`.