Question

a ball is thrown vertically upwards from ground with an initial velocity of 48 feet per second. the position function is given by
s(t)= -16t2+48t
(a) after how many seconds does the ball attains its maximum height?
(b) what is the maximum height the ball attains?
(c) after how many second does the ball return to the ground?

Answer 1

a)1.5 sec
b)36 m
c)3 sec

Answer 2

positiion(x)-16t^2+48
Velocity as the maximum must be zero
Velcotiy(x) -32t+48
set it = to zero

Answer 3

The line of the equation is a parabola, and the maximum height of the ball will be the highest point on the graph (vertex) which is on the axis of symmetry. The expression to find the axis of symmetry is\[-b \div2a\] so \[-48\div2\left( -16 \right) = 1.5\] and that gives you the x-value (time) when the ball will be at its maximum height. Now plug that into the position function: \[s(t)=-16(1.5)^{2}+48(1.5)\]\[s(t)=-36+72\]\[s(t)=36\] so 36 feet is the maximum height
For the time that the ball reaches the ground again, just multiply the value of the axis of symmetry by two and you get 3.