A rocket is launched from ground level with an initial vertical velocity (v0) of 176 ft/s. After how many seconds will the rocket hit the ground? (Hint: h(t) = -16t2 + v0t + h0).

Question

ash2326 · Answer

@tomtom777 You have the equation for the height of the rocket at any time t
$$h(t)=-16t^2+v_0 t+h_0$$
$$h_0= \text{height of the launch pad. Here it's 0, since rocket is launched from ground}$$
$$v_0=\text{initial vertical velocity}$$

ash2326 · Answer

Do you understand this?

anonymous · Answer

no

anonymous · Answer

Taking the initial height h0 =0 you get:$$h(t)=-16t^2+v_{0}t$$Now, since we are calling ground level h=0, we need to find t such that h(t)=0.  We have:$$-16t^2+v_{0}t=0$$We are given the initial velocity:$$-16t^2+176t=0$$We can factor out a t and a factor of 16:$$16t(-t+11)=0$$We solve each factor separately.  This equation is true is 16t=0 (i.e., t=0) and/or if -t+16=0 (i.e., t=16).  Therefore, the rocket is at ground level at t=0 and at t=16 seconds :)

anonymous · Answer

oooops,  should say -t+11=0, giving t=11 seconds.  typo

anonymous · Answer

The rocket hits the ground after 11 seconds.