quadratic equation

Question

One day Pokémon trainer Ash saw a rare Pokémon Mew on the ground in front of him. To catch the Pokémon he had to throw the Pokéball on it. The trajectory of the ball follows a parabola. In particular, the height of the ball above the ground, ℎ (in metres), at time 𝑡 (in seconds) after throwing the ball is given by: ℎ = −2𝑡2 +3𝑡 +2

 

(a) At which second will the ball reach the maximum height? 

(b) What is the maximum height the ball will reach?

(c) Calculate the time required for the ball to hit the ground. (Hint: what is the height of the ball when it hits the ground?)

(d) Suppose Mew is able to go away in 3 seconds, can Ash catch it?

Answer 1

ℎ = −2𝑡² +3𝑡 +2

 

The vertex of the parabola will be at maximum height

since the parabola opens downward.

Vertex is at (-b/2a, h(-b/2a))

 

-b/2a = -3/(2(-2)) = 3/4 seconds to reach maximum height

 

h(3/4) = -2(3/4)² + 3(3/4) + 2 = -2(9/16)+ 9/4 +2

          = -9/8 + 18/8 + 2 = 3 1/8 m or 3.125 m maximum height

 

At the point where the ball reaches the ground the height is zero

−2𝑡² +3𝑡 +2 = 0

 

t = [-2 ±√((3)²-4(-2)(2))]/(2(-2))

t = [-2 ±√(9+16)]/(-4)

t = (-2 ± √25)/(-4)

t = (-2 ± 5)/(-4)

t = -3/4 or 7/4

 

Discarding the negative it will take 1 3/4 seconds to reach the ground

 

Yes Ash can catch Mew