jean threw a disc in the air. the height of the disc can be modelled by the function h(t)=-5t^2+31.5t+2 what h(t) is height in meters after t seconds. patrick fired a paintball at the disc. the path of the paintball is modelled by the function g(t)=30t+1 with the same units. at what time(t) and how high(h) of the ground will the disc be when it is hit?
To find the time and height at which the disc is hit by the paintball, we need to set the two functions h(t) and g(t) equal to each other and solve for t.
Setting h(t) = g(t), we have:
-5t^2 + 31.5t + 2 = 30t + 1
Rearranging the equation and combining like terms, we have:
-5t^2 + 1.5t + 1 = 0
We can solve this quadratic equation by factoring or by using the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = -5, b = 1.5, and c = 1.
Plugging in these values, we get:
t = (-1.5 ± √(1.5^2 - 4(-5)(1))) / (2(-5))
= (-1.5 ± √(2.25 + 20)) / (-10)
= (-1.5 ± √22.25) / (-10)
Simplifying further, we have:
t ≈ (-1.5 + 4.71) / (-10) or t ≈ (-1.5 - 4.71) / (-10)
≈ 0.321 or t ≈ 0.671
We discard the negative value since time cannot be negative in this context.
Therefore, the disc is hit after approximately 0.321 seconds.
To find the height of the disc at this time, we substitute t = 0.321 into the h(t) function:
h(0.321) = -5(0.321)^2 + 31.5(0.321) + 2
Simplifying, we get:
h(0.321) ≈ 8.08
Therefore, the disc is approximately 8.08 meters above the ground when it is hit.