h=-16t^2+30t+7?

Question

Help I am lost!

The height h in feet of an object after t seconds is given by the function h=-16t^2+30t+7
How long will it take the object to hit the ground? Round your answer to the nearest thousandth, and explain which solution is the correct one.

Answer 1

The equation is of the form

x(t) = x0 + v0 * t + ( a * t ^2 ) /2

Since the t^2 term has the coefficient -16, the acceleration is -32 feet per second squared, or 1 g. The term in t has coefficient 30, which means the initial velocity is 30 fps upwards. The constant is the initial elevation, 7 feet.

The time required will be the time to accelerate to zero plus the time to accelerate from that height back to zero.

Time to reach zero vertical speed:
∂v = 30 fps
t = ∂v / a = 30/32 = 15/16 seconds to reach apogee
How high is apogee?
x(15/16) = 7 + 30(15/16) - 16(15/16)^2 = 49.1875 ft

Now we have to drop that distance:
x = x0 + v0*t + a*t^2/2 but v0 = 0 (at apogee):
0 = 49.1875 - 16*t^2
t^2 = 49.1875 / 16 = 3.07421875
t = sqrt(3.07421875) = 1.7533450173881922 seconds

The total time is 1.7533450173881922 + 15/16 = 2.6908 seconds. There you go.

Answer 2

In this problem, the object is initially 7 ft above the ground, and is tossed upwards at 30 ft.sec initial velocity. So we solve for for h=0. The solution will involve a quad. eqtn, and the negative t answer has to be discarded.

Answer 3

What can you say is true about h when the object hits the ground?

Answer: h = 0.

So set h = 0 and solve for t. You can use quadratic formula or completing the square.

Answer 4

You are trying to find the time when h=0 (the height when it hits the ground).

0=-16t^2+30t+7

The rest is just algebra. I would use the quadratic formula.

[-b +/- root (b^2 -4ac)] / 2a