A projectile is launched from ground level with an intital velocity of vo feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t^2+v0^t. Find the time(s) that the projectile will (a) reach a height of 128ft and (b) return to the ground when v0=96 feet per second.
(a)find the time(s) that the projectile will reach a height of 128ft when v0=96 feet per second. Select the correct choice below and if necessary fill in the answer box to complete your choice.

Question

A projectile is launched from ground level with an intital velocity of vo feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t^2+v0^t. Find the time(s) that the projectile will (a) reach a height of 128ft and (b) return to the ground when v0=96 feet per second. 
(a)find the time(s) that the projectile will reach a height of 128ft when v0=96 feet per second. Select the correct choice below and if necessary fill in the answer box to complete your choice.

phoenixfire · Answer

You have your equation which gives height based on time. 
$$s=-16t^2+v_{0}^t$$
What you need to do is rearrange that equation for t, plug in the known values and then you'll have your time. That's for (a). $$v_{0}=96$$$$s=128$$

phoenixfire · Answer

Is v to the power of t. or v times t?
$$v_{o}^t$$$$v_{o}t$$

anonymous · Answer

the second one v0t

anonymous · Answer

$$v _{0}t$$

phoenixfire · Answer

Okay. So you're set? You know how to do it now?

anonymous · Answer

I dont understand how to get answer b I know answer a is 2,4

phoenixfire · Answer

Would you not use your rearranged equation for t and have your final position s=0.

anonymous · Answer

i did 0=-16t^2+96t but that is not right so i dont know

phoenixfire · Answer

$$0=-16t^2+96t$$$$16t^2=96t$$$${16t^2 \over t}={96t \over t}$$$$16t=96$$$$t={96 \over 16}$$$$t=6$$
Why is that wrong?

anonymous · Answer

that is right i put 0,6 as the answer and just 6 is the answer sorry