What is the initial velocity of a golf ball?

Agolf ball is hit off a tee with an initial velocity of 128 feet per second. If h(t)=-16t^2 +128t is the equation that models this event,



a)after how many seconds will the the golf ball reach a height of exactly 112 feet?

b)after how many seconds will the ball reach its maximum height?

c)what is the maximum height of the golf ball?

d)after how many seconds will the ball touch the ground?



PLEASE HELP!!!! i have NO IDEA how to do this. explain step by step.What is the initial velocity of a golf ball?a) -16t^2 +128t =112

-t^2 + 8t - 7 = 0

t^2 - 8t + 7 = 0

(t - 1)(t - 7) = 0

t = 1 , 7 =%26gt; at 1 sec and 7 sec in the way up and down respectively



b) a = -16 %26lt; 0 , the parabola opens down which makes the vertex at the maximum hence:

t = -b/2a = -128/-32 = 4 sec =%26gt; time to reach the max h.



c) h(4) = 256 feet max h



d) t^2 - 8t = 0

t(t - 8) - 0

t = 0, 8 =%26gt; hence after 8 seconds it will hit the ground.What is the initial velocity of a golf ball?a) is simple. Plug 112 feet for h(t) and solve for t. 112 = -16t^2 + 128t. I'll let you solve this yourself.


For b you need to differentiate the h(t) function.
b) h'(t) = -32t + 128 = 0
-128 = -32t
t = 4

Second derivative test:
h''(t) = -32 (negative means concave down --%26gt; maximum height)

so at t = 4, the ball is at its max height.

c) now you know the time of its max height, plug t=4 back to its original function for height.
Solve h(4) yourself.

d) When it reaches the ground, its height is 0. Solve for h(0). You'll get two answer, one of them is 0 where it first starts. The other will be 0 = -16t^2 + 128t
Finding t is very simple.What is the initial velocity of a golf ball?appears golf ball thrown up... from the model equation.



initial velocity is 128 ft/sec



h(t) = ut + 1/2 at^2.....112 = -16t^2 + 128t .....t^2 -8t +7 =0.........(t -7) ( t-1) =0



t =1,7



but v = u +at.............v(t) = 128 -32 t ...a nd at max height v(t) =0 ....so, t =4 sec for max height



so, after 1 sec going up, 7sec coming down for 112 feet.



If h(t)=-16t^2 +128t substitute t = 4sec ....so, max hight is 256 feet



after 4 sec equation is h(t) = 16t^2 and distance is 256 feet so, 4sec after reaching hmax and 8 sec after thrown up.