Equations of trajectory of a projectile are x=4 t, `y= 3t - 5 t ^(2) ` , where x and y are in metre and t is in second . Find (i) maximum height attained by the projectille , (ii) horizontal range of the projectile .

Explanation

` x= 4t , y= 3t - 5 t^(2)`
` therefore`vertical velocity `= ( dy) /( dt) =3-10 t `
which is zero at the maximum height
`therefore 3-10 t=0 or , t=(3)/(10) s`
(i) maximum light
` H= 3 xx(3)/(10) - 5 xx((3)/(10 ))^(2)= 0.45`m
(ii) time taken to reach the maximum height `=(3)/(10) s ` , and so the time taken to reach the ground from maximum height `=(3)/(10) s.` Therefore , time of flight `=(3)/(10) +(3)/(10) =(3)/(s) s.`
`therefore ` horizontal range , ` R = 4 xx3/5 = 2.4 m`