Question

A ball is thrown into the air with an upward velocity of 36 ft/s. Its height h in feet after t seconds is given by the function h = –16t2 + 36t + 10. a. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. b. What is the ball’s maximum height? 2.25 s; 10 ft 1.13 s; 30.25 ft 1.13 s; 70.75 ft 1.13 s; 32.5 ft

Answer 1

We can do this by converting the equation to vertex form:-

h = -16t^2 + 36t + 10

= -16(t^2 - 2.25t) + 10

= -16 [ (t  - 1.125)^2 - (1.125)^2] + 10

= -16(t - 1.125)^2 + 20.25 + 10

= -16(t - 1.125)^2 + 30.25  

So the answer is 1.13s and 30.25 ft.

=  (

Answer 2

Answer:

(a) 1.125s ~ 1.13s (b) 30.25 ft at 1.13s

Step-by-step explanation:

(a) The motion described by the ball corresponds to accelerated motion, which is analyzed in physics from free fall in Newtonian mechanics. Nonetheless, this motion can be analyzed from an algebraic point of view because the motion is determined by a quadratic expression and its properties provide information of intercepts at x and y-axis and its vertex. For our purposes, the problem will be tackled from algebraic point of view. Firstly, it is briefly necessary to remind that any quadratic equation has some coefficients; namely three coefficients as follows

$ax^2+bx+c$

Nonetheless, this equation is not sufficient to say something about the problem itself, then it is required to relate the previous equation with the motion of the ball and that is done using the criteria of a function. If the balls is moving upward, then its position is changing in time, i.e, change of height is dependent of the time running. With such criteria, the equation can be written as;

$h(t)=-16t^2+36t+10$

This is a quadratic function in which the maximum height is calculated from its vertex. Avoiding any misunderstanding about the used variables, t represents the x-variable and h represents the y-variable Using the vertex criteria and considering the values of the mentioned coefficients a=-16, b=36, c=10, the time in which the maximum height is reached is given by,

$x=-\frac{b^2}{2a}=-(\frac{36}{2(-16)})s=(\frac{9}{8})s=1.125s = 1.13s$

(b)  Given the time in which the ball reached its maximum height, the height can be determined from the mentioned function and substituding t=1.13 or t=9/8 into the function, hence

$h(t)=-16t^2+36+10$

$h(t=\frac{9}{8})=-16(\frac{9}{8})^2+36(\frac{9}{8})+10\\\\h(t=\frac{9}{8})=-16(\frac{81}{64})+36(\frac{9}{8})+10= 30.25 ft$