OR Note that Now the maximum value that sin can be is 1; this will happen if is 900, or = 450. 2β 2β β And that’s it. However the expression for the range always be in the form above, so you need to be familiar with the won’t differentiation method. 1981 (a) Establish an expression, in terms of initial speed and angle of inclination to the horizontal for the range u α, of a projectile on a horizontal plane through the point of projection. Deduce that the maximum range for a given is u u2/g. 1989 A particle is projected with speed at an angle to the horizontal. u α The range of the particle on the horizontal plane through the point of projection is R. Show that is a maximum when = 450. R α If = , find the two possible values of R α. the ratio of the greatest height to the range is 2:5, find α. Trickier example Given the expression for the range in the first line below, we can find the value of that would result in ϑ maximum range as follows:
OR Note that Now the maximum value that sin 2β can be is 1; this will happen if 2β is 900, or β = 450. And that’s it. However the expression for the range won’t always be in the form above, so you need to be familiar with the differentiation method. 1981 (a) Establish an expression, in terms of initial speed u and angle of inclination to the horizontal α, for the range of a projectile on a horizontal plane through the point of projection. Deduce that the maximum range for a given u is u2/g. 1989 A particle is projected with speed u at an angle α to the horizontal. The range of the particle on the horizontal plane through the point of projection is R. Show that R is a maximum when α = 450. If R = , find the two possible values of α. the ratio of the greatest height to the range is 2:5, find α. Trickier example Given the expression for the range in the first line below, we can find the value of ϑ that would result in maximum range as follows:
3.%20Projectiles.docx