An object is projected vertically upward with an initial velocity of v0 ft/sec, and its distance s(t) in feet above the ground after t seconds is given by....

Question

the formula s(t) = −16t2 + v0t.

 

(a) If the object hits the ground after 12 seconds, find its initial velocity v0.
v0 =        ft/sec

(b) Find its maximum distance above the ground.
s(t) =        ft

 

I am unsure how to approach this problem. I feel as if some information is missing or am I just lacking some generic knowledge used in the formula for velocity?

Answer 1

a) First of all, when something is thrown up and returns to the ground the total displacement of the object is 0 (zero)

 

So s(t) = 0 ft

 

0 = -16*12² + v₀*12

 

0 = -2304 + v₀*12

 

2304 = Vo*12

 

2304/12 = Vo

 

192 ft/sec = Vo

 

b) Next  when something is thrown up and returns to the ground the time it takes going up = the time it takes falling down. Since the total time was 12 sec, the time going up is 6 sec. That is the time to reach its highest point.

 

s(t) = -16*6² + 192*6

 

s(t) = 576 ft

 

Hope this helps!