Application of quadratic models If air resistance is neglected, the height h (in

Question

Application of quadratic models If air resistance is neglected, the height h (in feet) of an object projected directed directly upward from an initial height ho feet with initial velocity Vo feet per second is h(t)-1612+ vot + ho where t is the number of seconds after the object is projected. The coefficient of f2 is a constant based on the gravitational force of Earth. A ball is projected directly upward from an initial height of 75 ft with an initial velocity of 112 ft per sec. a) Give the function that describes the height of the ball in terms of time t. b) After how many seconds does the ball reach its maximum height? What is the maximum height? c) For what interval of times is the height of the ball greater than 200 ft? d) After how many seconds will the ball hit the ground?

Explanation / Answer

The height h(t) of an object ( in feet), projected directly upwards from an initial height h0 ft. with initial velocity v0 is given by h(t) = -16t2 +v0t +h0.

a). Here, h0 = 75 ft. and v0 = 112ft. per second. Hence, the required function is h(t) = -16t2 +112t +75.

b). The function h(t) = -16t2 +112t +75 represents a downwards opening parabola. Also, h(t) = -16(t2 -112t) = -16(t2 -2*56t +562)+16*562 = -16(h-56)2+ 50176. This is the vertex form of the parabola with vertex at (56, 50176). Since the vertex is the highest point of a downwards opening parabola, hence the maximum height is 50176 ft. Also, when h(t) = 50176, then t = 56. Thus, the maximum height is reached after 56 seconds.

c). If h(t) > 200, then -16t2 +112t +75 > 200 or, -16t2 +112t > 125 or, 16t2 -112t < 125 or, t2 -7t <125/16 or, t2 -2*7t/2 + (7/2)2 < 125/16 + (7/2)2 or, (t-7/2)2 < 321/16 so that t-7/2 < ?(321/16) or, t-7/2 < 4.479 (approx.) so that t < 4.479 +3.5 = 7.98 seconds ( on rounding off to 2 decimal places). Thus, for a period of 7.98 seconds(approx..), the height of the ball is greater than 200 ft.

d).When the ball hits the ground, h(t)=0 so that -16t2 +112t +75=0 or, 16t2-112t-75=0. On using the quadratic formula, t =[112±?{(-112)2-4*16*(-75)}]/2*9= [112±?(12544+4800)]/18 =(112±?17344)/18 = (112±131.70)/18 = (112+131.70)/18 ( as t cannot be negative) = 13.54 seconds( on rounding off to 2 decimal places).Thus, the ball will hit the ground after 13.54 seconds(approx.),

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