A projectile is launched from a platform 20 feet high with an initial velocity of 48 feet per second, The height h of the projectile at t seconds after launch is given by h = –16t^2 + 48t + 20 feet.
(a) How many seconds after launch does the projectile attain maximum height? (b) What is the maximum height
well, if you are not allowed to use calculus you must complete the square to find the vertex of that parabola.
16 t^2 - 48 t -20 = -h
t^2 - 3 t - 1.25 = -h/16
t^2 - 3 t = -h/16 + 1.25
t^2 - 3 t + 9/4 = -h/16 + 3.5
(t - 1.5)^2 = - 1/16 ( h - 56)
so at peak t = 1.5 and h = 56
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check with calculus
when h is max, dh/dt = 0
0 = -32 t + 48
so t = 48/32 = 1.5 sure enough
then h = -16(2.25) + 48(1.5) + 20
= 56 ok
To find the answers to these questions, we need to understand the concept of maximum height for a projectile motion and apply some basic calculus principles.
(a) How many seconds after launch does the projectile attain maximum height?
To find the time at which the projectile reaches its maximum height, we need to determine the value of 't' that gives us the maximum value of the height function.
The height of the projectile is given by the equation h = -16t^2 + 48t + 20.
To find the maximum height, we need to find the vertex of the parabolic function. The vertex is the point where the function reaches either its highest or lowest point.
The x-coordinate of the vertex can be found using the formula:
x = -b / (2a),
where 'a' and 'b' are the coefficients of the quadratic equation.
In this case, the quadratic equation is -16t^2 + 48t + 20,
so, 'a' = -16 and 'b' = 48.
Now inserting these values into the formula, we get:
t = -48 / (2 * (-16)).
= -48 / -32.
= 1.5.
Therefore, the projectile attains its maximum height 1.5 seconds after launch.
(b) What is the maximum height?
To find the maximum height, substitute the value of 't' (1.5 seconds) into the height function:
h = -16(1.5)^2 + 48(1.5) + 20.
= -16(2.25) + 72 + 20.
= -36 + 72 + 20.
= 56 feet.
Hence, the maximum height attained by the projectile is 56 feet.