1. ## projectiles

A stone is projected in a direction which makes an angle of 45° above the horizontal. It strikes a small target whose horizontal and vertical distances from the point of projection are 120m and 41.6m respectively. The target is above the level of the point of projection.

a) Find the speed of projection and show that the time taken for the stone to reach the target is 4s.

b) Determine, correct to 2d.p., that the speed and direction of motion of the stone as it hits the target.

Ok i have been taught very little on projectiles.

I have these to play around with (column notation)

when t=0

a =
0
-g

or
(0i-gj)

V=
vcosα
-gt+vsinα

or
((vcosα)i + (-gt+vsinα)j)

r=
(vcosα)t
(-gt²)/t + (vsinα)t

or
(((vcosα)t)i + ((-gt²)/t + (vsinα)t)j)

not really sure how to start the question

2. Hello djmccabie

You must know that horizontal speed remains constant(if air resistance is ignored)
Vertical velocity$\displaystyle =v sin(45)$
Horizontal velocity$\displaystyle =v cos(45)$

Horizontal distance$\displaystyle =120=v cos(45).t$(t=time period)

Vertical distance$\displaystyle =41.6=$$\displaystyle v sin(45).t-\frac{1}{2}(9.81)(t^2)$
($\displaystyle s=ut+\frac{1}{2}at^2$)

Solve both equations to get v&t

b) Vertical speed$\displaystyle =v sin(45).t-9.81t=v sin(45).4-9.81(4)$
Horizontal speed$\displaystyle =v cos(45)$
Resultant vel. at that point$\displaystyle =\sqrt{(vel.vertical)^2+(vel.horizontal)^2}$

Direction of motion$\displaystyle =arctan(\frac{vel.vertical}{vel.horizontal})$ angle to the horizontal

Hope this helps