Asked by Alex
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)
Answers
Answered by
Henry
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.
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.
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