A projectile is shot vertically into the air. Its height, h, in metres, after t seconds is approximately modelled by the relation h=-5t^2+120t+125:
c) For how many seconds is the projectile in the air? (1 mark)
According to your equation, the object was 125 m above ground (t=0) , when it was shot
solve for h = 0 , this will give you the two times it is on the ground
-5t^2 + 120t + 125 = 0
divide by -5
t^2 - 24t - 25 = 0
(t-25)(t+1) = 0
t = -1 or t = 25
but, this only makes sense for t≥ 0, so the object was in the air from 0 to 25 seconds
or 25 seconds
To determine how long the projectile is in the air, we need to find the time when the height is zero.
Given the equation for the height of the projectile: h = -5t^2 + 120t + 125
Set h = 0 and solve for t:
0 = -5t^2 + 120t + 125
This is a quadratic equation. We can solve it by factoring or by using the quadratic formula.
Let's use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = -5, b = 120, and c = 125.
Plugging in these values into the quadratic formula:
t = (-120 ± √(120^2 - 4(-5)(125))) / (2(-5))
Simplifying:
t = (-120 ± √(14400 + 2500)) / -10
t = (-120 ± √(16900)) / -10
Now, we can solve for t by finding the values of t that satisfy the equation:
t = (-120 + √16900) / -10
t = (-120 + 130) / -10
t = 10 / -10
t = -1
and
t = (-120 - √16900) / -10
t = (-120 - 130) / -10
t = -250 / -10
t = 25
Since time cannot be negative in this context, we disregard the negative solution.
So, the projectile is in the air for 25 seconds.
To find the number of seconds the projectile is in the air, we need to determine the time when the height, h, becomes zero.
The given equation for the height of the projectile is h = -5t^2 + 120t + 125.
To find the time when the height is zero, we set h = 0 and solve for t:
0 = -5t^2 + 120t + 125
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -5, b = 120, and c = 125.
t = (-120 ± √(120^2 - 4(-5)(125))) / (2(-5))
Simplifying further:
t = (-120 ± √(14400 + 2500)) / (-10)
t = (-120 ± √(16900)) / (-10)
t = (-120 ± 130) / (-10)
Now we have two possible solutions:
1) t = (-120 + 130) / (-10) = 10 / (-10) = -1
2) t = (-120 - 130) / (-10) = -250 / (-10) = 25
Since time can only be positive, we disregard the negative solution (-1).
Therefore, the projectile is in the air for 25 seconds.