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
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?
2 answers
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
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