A baseball player hits a high pop-up with an initial upward velocity of 30 m/s, 1.4 m above the ground. The height, h, in metres, of the ball t seconds after being hit is modelled by the function h(t) = -4t^2+30t+1.4. How long does a player on the opposing team have to get under the ball if she catches it 1.4 m above the ground? (2 marks)
h(t) = -4t^2 + 30t + 1.4
Tr = -B/2A = -30/-8 = 3.75 s. = Rise
time.
Tf = Tr = 3.75 s. = Fall time.
Tr+Tf = 3.75 + 3.75 = 7.5 s. To get
under the ball.
To find how long the opposing player has to get under the ball, we need to find the time when the height of the ball is 1.4 meters.
Given the model function h(t) = -4t^2 + 30t + 1.4, we can set h(t) equal to 1.4 and solve for t.
-4t^2 + 30t + 1.4 = 1.4
Simplifying the equation:
-4t^2 + 30t = 0
Adding 4t^2 - 30t to both sides:
4t^2 - 30t = 0
Factoring out a common factor:
2t(2t - 15) = 0
Setting each factor equal to zero:
2t = 0 or 2t - 15 = 0
Solving for t in each case:
2t = 0 --> t = 0 (Ignore, as a player cannot catch the ball when it's just hit)
2t - 15 = 0 --> t = 7.5
Therefore, the opposing player has to get under the ball within 7.5 seconds to catch it 1.4 meters above the ground.
To find out how long a player on the opposing team has to get under the ball if she catches it 1.4 m above the ground, we need to determine the time at which the height of the ball is equal to 1.4 m above the ground.
Given the function h(t) = -4t^2 + 30t + 1.4, where h(t) represents the height of the ball at time t:
1. Set h(t) equal to 1.4 m and solve for t:
-4t^2 + 30t + 1.4 = 1.4
2. Simplify the equation:
-4t^2 + 30t = 0
3. Move all the terms to one side to form a quadratic equation:
-4t^2 + 30t = 0
4. Factor out the common term:
t(-4t + 30) = 0
5. Set each factor equal to zero and solve for t:
t = 0 or -4t + 30 = 0
6. Solve the second equation:
-4t + 30 = 0
4t = 30
t = 30/4
t = 7.5
Therefore, the player on the opposing team has 7.5 seconds to get under the ball if she catches it 1.4 m above the ground.