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Question

The path of a ball being thrown straight up in the air is given by

y = −5t^2 + 14t + 3 where t is the time in seconds that the ball is in the air and y is the height

(in meters). What is the maximum height of the ball?

Answer 1

This is a parabola which is expressed as at²+bt+c. For this case:

a=-5

b=14

c=3

The vertex of any parabola can be found using t = -b/(2a).

In this case, the vertex is at t=-14/(-10) = 1.4

y(1.4)=-5(1.4)²+14(1.4)+3 = -9.8 + 19.6 + 3 = 12.8 meters

Answer 2

The path of a ball being thrown straight up in the air is given by

y = −5t^2 + 14t + 3 where t is the time in seconds that the ball is in the air and y is the height

(in meters). What is the maximum height of the ball?

For y to be maximum derivate of y with respect is zero

so d/dt (y )= 0

-10t +14 = 0

10t= 14

t= 14/10 = 7/5 = 1.4 s

Plug in t= 1.4 in above equation

y = -5t^2 +14t+3

= -5*1.4*1.4 +14*1.4 +3

=12.8 m