Find t with the equation h=-2.7t^2+13.5t+14

Question

American Astronauts working on a space station on the moon toss a ball into the air. The height of the ball is represented by the equation h=2.7t^2+13.5t+14, where t represents the number of seconds since the ball was thrown, and h represents the height of the ball in get.

a) To the nearest hundredth of a second, after how much time does the ball hit the ground

b) to the nearest tenth of a for, what is the greatest height that the ball achieves?

c) At what time (to the nearest hundredth of a second) did the ball reach its maximum height?

Answer 1

As Timothy stated, the sign of the t² term is incorrect in the problem statement, although it is correct in the title.  The correct equation is

 

h(t) = -2.7t² + 13.5t + 14

 

(a)

 

Set h(t) = 0 and solve the resulting quadratic equation for t.  You will get a negative and a positive solution.  Discard the negative solution, since t starts at t = 0.

 

(b)

 

h_max occurs at the vertex of the h vs. t parabola, which is at time

 

t = -13.5/[2(-2.7)] s = ? s

 

Evaluate h(t) at this time to get h_max.

 

(c)

 

It is the time found in part (b).

 

 

Answer 2

I think you wrote the equation wrong. The first coefficient should be a negative value in order to describe a parabola tracing a tossed ball.