I need math help?

Question

The equation h= -t to the second power plus 112t to the second power gives the height of an arrow, shot upward from the ground with an initial velocity of 112ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to rach a height of 180ft.

Answer 1

I think your equation should read:

h=-t^2 + 112t

to find out when it reaches 180ft put h = 180

ie

180 = -t^2 + 112t

using the quadratic formula gives the answers 1.63 secs (on the way up) and 110.37 secs (on the way down).

Incidently the greatest height the arrow reaches is after 56 secs and is 3136 ft.

If my assumption of your equation is correct this represents the flight of an arrow under approx 1/5 gravity of earth. Think of it as an arrow shot on the moon.

Answer 2

I think you have missed to put some variable in the expression. Because

h = - t^2 + 112t^2 (there is no variable for velocity here)
h = 111t^2 eq. (1)

so h= 111t^2 and this is a curved path the arrow is following.

So by the amount of information you have given t=? when h=180ft.

Using eq. (1)
180 =111 t^2
t^2 = 180/111
t^2 = 1.6216
t = 1.2734 in seconds that is

But there is no relationship with the velocity you have given in the data. So most probably your initial expression has something missing.

Hope this helps.

Answer 3

They give you an expression for height in terms of time. So set the height equal to 180 and solve for t. You might get to different answers, because the arrow hits a certain height on the way up and hits the same height on the way back down.

Answer 4

Are you sure you have that right - both ts are to the second power ? Doesn't sound right to me.