Question

A baseball thrown upward at initial speed of 9.5m/s.
The approximate height of the ball, h metres, after t seconds is given by the equation
h = -4.9t2 + 9.5t+1.
At the same time, a golf ball is hit upward at an initial speed of 12m/s. The approximate height of the ball, h metres, after t seconds is given by the equation h=-4.9t2 + 12t

a) When are both balls at the same height?

b) What is this height?​

Answer 1

Answer:

a) 0.4 s; b) 4.016 m

Step-by-step explanation:

a) Time to same height

$\begin{array}{lrcl}(1) & h &=& -4.9t^{2} + 9.5t + 1\\(2) & h & = & -4.9t^{2} + 12t\\& 0 & = & 2.5t - 1\\& 1 & = & 2.5t\\& t & = & \dfrac{1}{2.5}\\\\& t & = & \textbf{0.4 s}\\\end{array}$

b) Height

Insert the value of t into one of the equations.

$h  = -4.9(0.4)^{2} + 12(0.4) = -4.9(0.16) +4.8 = -0.784 + 4.8 = \textbf{4.016 m}$

The diagram below shows that the trajectories of the balls intersect after 0.4 s at a height of 4.016 m.