1. ## Projectiles

A stone is projected in a direction which makes an angle of 45 degrees above the horizontal. It strikes a small target whose horizontal and vertical distances from the point of projection are 120 m and 41.6 m 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. [8 marks]

b) Determine, correct to 2 decimal places, the speed and direction of motion of the stone as it hits the target. [7 marks]

OK my tutor has taught me some equations but I have not had time to properly learn how to answer the questions yet. The equations are in column format representing x and y.

a (acceleration) =
0
-9.8

v (velocity) =
Vcosα (+C when not at origin)
Vsinα - 9.8t (+C when not at origin)

r (Displacement) =
Vcosα(t) (+C)
Vsinα(t) - (1/2)(gtē) (+C)

iv put the numbers in for V and α in each equation but i'm not sure what to do with them. help please?

2. Originally Posted by djmccabie
A stone is projected in a direction which makes an angle of 45 degrees above the horizontal. It strikes a small target whose horizontal and vertical distances from the point of projection are 120 m and 41.6 m 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. [8 marks]

b) Determine, correct to 2 decimal places, the speed and direction of motion of the stone as it hits the target. [7 marks]

OK my tutor has taught me some equations but I have not had time to properly learn how to answer the questions yet. The equations are in column format representing x and y.

a (acceleration) =
0
-9.8

v (velocity) =
Vcosα (+C when not at origin)
Vsinα - 9.8t (+C when not at origin)

r (Displacement) =
Vcosα(t) (+C)
Vsinα(t) - (1/2)(gtē) (+C)

iv put the numbers in for V and α in each equation but i'm not sure what to do with them. help please?
$\Delta x = v_x \cdot t$

$120 = v\cos(45) \cdot t$

$\Delta y = v_{y0} \cdot t - \frac{1}{2}gt^2$

$41.6 = v\sin(45) \cdot t - 4.9t^2$

since $\cos(45) = \sin(45)$ ...

$41.6 = 120 - 4.9t^2$

solve for $t$ ... then solve for the value of the initial velocity.

for the second part ... $v_x$ remains constant

$v_y = v_{y0} - gt$

speed = $\sqrt{v_x^2 + v_y^2}$

direction relative to the horizontal = $\arctan\left(\frac{v_y}{v_x}\right)$