Answer to Question #177003 in Differential Equations for daniel wagaye

Question

An object is thrown vertically upward from a height of h0 ft with an initial speed of v0 ft/sec. Its height h (in feet) after t seconds is given by

h= -16t² + h0. Given this, if it is thrown vertically upward from the ground with an initial speed of 64 ft/sec,

(a) At what time will the height of the ball be 15 ft? (two answers)

(b) How long will it take for the ball to reach 63 ft?

Answer 1

The correct formula should show the dependence not only on time, but also on the initial velocity:

$h = -\dfrac{gt^2}{2}+v_0t+h_0 = -16t^2+v_0t+h_0.$ The formula from the problem is correct for the motion without initial velocity. If the object is thrown with an initial speed of 64 ft/sec, the height is given by

$h= -16t² + 64t + h_0.$

(a) $15= -16t² + 64t +0.$

$16t^2-64t+15 = 0.$

$D = 64^2-4\cdot16\cdot15 = 3136 = 56^2.$

$t_{1,2} = \dfrac{64\pm56}{2\cdot16} = 0.25\; \text{or} \; 3.75.$

(b) $63= -16t² + 64t +0.$

$16t^2-64t+63 = 0.$

$D = 64^2-4\cdot16\cdot63 = 64 = 8^2.$

$t_{1,2} = \dfrac{64\pm8}{2\cdot16} = 1.75\; \text{or} \; 2.25.$

So the time to reach 63 ft is 1.75 seconds.