Question

An archer shoots an arrow from ground level towards a tower of a castle 45m high. The height of the arrow above the ground after t seconds is given by the equation h = 30t - 5t²

a) after how much time does the arrow reach the top of the tower ?

b) If the arrow doesn't hit anything, after how long will it he at ground level again?

Answer 1

h = 30t - 5t² = 45
5 t^2 -30 t + 45 = 0
t = 3 s
when does h = 0?
5 t^2 - 30 t = 0
t (30- 5 t) = 0
t = 6 s

Answer 2

a)

When the arrow reach the top of the tower h = 45 m

h = 30 t - 5 t²

45 = - 5 t² + 30 t

Subtract 45 to both sides

0 = - 5 t² + 30 t - 45

Divide both sides by - 5

0 = t² - 6 t + 9

Now you must solve equation:

t² - 6 t + 9 = 0

The solution is t = 3

b)

If the arrow doesn't hit anything the arrow will fall to the ground and the height will be zero.

So slove:

30 t - 5 t² = 0

Divide both sides by - 5 t

- 6 + t =0

Add 6 to both sides

t = 6