In this question use $g = 9.8$ m s $^{-2}$ .

A particle $P\,$ is projected from a point $O\,$ with speed $U\,$ m s $^{-1}$ at an angle $\theta\,$ above the horizontal. Air resistance is modelled as negligible. When $P\,$ has moved a horizontal distance $x\,$ metres, its height above $O\,$ is $y\,$ metres.

By writing down expressions for $x\,$ and for $y\,$ in terms of the time $t$ , and eliminating $t$ , show that $x\tan\theta - \dfrac{gx^2}{2U^2\cos^2\theta}$ .

It is given that $θ=45°\theta = 45°$ , and that $y = 2$ when $x = 6$ . Find the speed of $P\,$ at the point where $x = 6$ and $y = 2$ .

In this question use $g = 9.8$ m s $^{-2}$ .

By writing down expressions for $x\,$ and for $y\,$ in terms of the time $t$ , and eliminating $t$ , show that $x\tan\theta - \dfrac{gx^2}{2U^2\cos^2\theta}$ .

It is given that $θ=45°\theta = 45°$ , and that $y = 2$ when $x = 6$ . Find the speed of $P\,$ at the point where $x = 6$ and $y = 2$ .

3.2 Kinematics

Question 42

Markscheme

3.2 Kinematics

Question 42

Markscheme

3.2 Kinematics

Question 42

3.2 Kinematics Questions

3.2 Kinematics

Question 42

3.2 Kinematics Questions