5.05 Projectile motion further examples

The three equations we have used for projectile motion are;
For horizontal displacement x = uxt (i)
For vertical displacement y = uyt - 1/2gt2   (ii)
For vertical velocity vy = uy - gt     (iii)

Equations (i) and (ii) are variations of suvat equation 3 (s = ut + 1/2at2)
Equation (iii) is a variation of suvat equation 2 (v = u - at)

In some examples of projectile motion the projectile is fired horizontally from a higher ground level (or the top of a building etc.)
Therefore in these examples the projectile's initial vertical velocity (uy) is zero
This means:
equation (ii) simplifies to y = 1/2gt2
equation (iii) simplifies to vy = gt

projectile question

Note the change in sign i.e. dropping the "-" from "- 1/2gt2 " and dropping the "-" from "- gt " and the change of direction of the y axis in the diagram above

In the previous example their was a change in direction of vertical component of the projectiles velocity (vy). Therefore we used "+" for up and "-" for down. Acceleration due to gravity (g) acts down so it would also be negative. y in the previous example represented the height above the ground.

In this example the vertical component of the velocity starts at 0 and then increases in magnitude but is always directed downwards. So as there is only one direction involved now then we just choose down to be "+"". This means g is now positive and y will now be the distance the object drops from the initial height it was launched from. This is illustrated by the change in the direction of the y axis in the diagram above.

A projectile is launched horizontally from the top of a building with an initial velocity of 10 ms-1
If the object lands 30 metres from the base of the building then what is the height of the building?

x = uxt    therefore t = x/ux    t = 30/10 = 3s
y = 1/2gt2    therefore y = 1/2 x 9.81 x 32 = 44.15m