Question

If an object is projected upward with an initial velocity of 48 ft pe second from a height h of 160 ft, then its height t second after it is projected is defined by the equation h= -16t 2 + 48t + 160. How many seconds after it is projected will it hit the ground?

Answer 1

You are looking for the time, t, when h(t) = 0, so... {{{h(t) = -16t^2+48t+160}}} Substitute h(t) = 0. {{{0 = -16t^2+48t+160}}} Rewrite in standard form: {{{-16t^2+48t+160 = 0}}} First factor -16. {{{-16(t^2-3t-10) = 0}}} Notice the change of sign in each term in the parentheses as a result of factoring the -16. Now apply the zero product rule. {{{t^2-3t-10 = 0}}} Solve by factoring. {{{(t+2)(t-5) = 0}}} Apply the zero product rule. {{{(t+2) = 0}}} or {{{(t-5) = 0}}} so then... {{{t = -2}}} or {{{t = 5}}} Discard the negative solution as the time, t, should be a positive value. {{{highlight(t = 5)}}}seconds.

Answer 2

If an object is projected upward with an initial velocity of 48 ft per second from a height h of 160 ft, then its height t second after it is projected is defined by the equation h= -16t 2 + 48t + 160. : How many seconds after it is projected will it hit the ground? ; When the object strikes the ground h = 0, therefore: -16t^2 + 48t + 160 = 0 : Simplify, divide by -16 t^2 - 3t - 10 = 0 Factor (t - 5)(t + 2) = 0 Positive solution t = 5 seconds to hit the ground : : Check solution in original equation h = -16t^2 + 48t + 160 h = -16(5^2) + 48(5) + 160 h = -400 + 240 + 160 h = 0