2016-04-11 08:23:36
a projectile is launched straight up with an initial velocity of 120 ft/s. if the acceleration due to gravity is -16 ft/s2, after about how many seconds will the object reach a height of 200 ft? h(t)= at 2 +vt+h0
2016-04-11 13:32:25

bearing in mind that the projectile is being launched from the ground, thus it has an initial height of 0. [latex]f qquad extit{initial velocity}\\ egin{array}{llll} qquad extit{in feet}\\ h(t) = -16t^2+v_ot+h_o end{array} quad egin{cases} v_o= extit{initial velocity of the object}\ qquad 120\ h_o= extit{initial height of the object}\ qquad 0\ h= extit{height of the object at "t" seconds}\ qquad 200 end{cases} \\\ 200=-16t^2+120t+0implies 25=-2t^2+15t \\\ 2t^2-15tr+25=0\\\ (2t-5)(t-5)=0implies egin{cases} 2t-5=0implies &t=frac{5}{2}\ t-5=0implies &t=5 end{cases}[/latex] notice the picture below, is more or less like so so it hits 200feet at two points, once on the way up, then again on the way down.