A projectile is fired from a canon at ground level with initial velocity `sqrt300` ms−1 at an angle of 30° to the horizontal.
The equations of motion are `(d^2x)/(dt^2) = 0` and `(d^2y)/(dt^2) = −10`.
- Show that `x = 15t`. (1 mark)
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- Show that `y = 5sqrt3t-5t^2`. (2 marks)
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- Hence find the Cartesian equation for the trajectory of the projectile. (1 mark)
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a.
`dotx = sqrt300\ cos30^@ = 10sqrt3 xx sqrt3/2 = 15\ text(ms)^(−1)`
`x = int 15\ dt = 15t + C_1`
`text(When)\ \ t = 0, \ x = 0 \ => \ C_1 = 0`
`:. x = 15t`
b. `doty = int −10\ dt = −10t + C_1`
`text(When)\ \ t = 0, \ doty = 10sqrt3 sin30^@ = 5sqrt3 \ => \ C_1 = 5sqrt3`
`:. doty = 5sqrt3-10t`
`y = int doty\ dt = 5sqrt3t-5t^2 + C_2`
`text(When)\ \ t = 0, \ y = 0 \ => \ C_2 = 0`
`:. y = 5sqrt3t-5t^2`
c. `x = 15t \ => \ t = x/15`
| `y` | `= 5sqrt3 xx x/15-5(x/15)^2` | |
| `y` | `=sqrt3/3 x-(x^2)/45` |