If an object is projected upward with an initial velocity of 112 feet per second, at what times will it reach a height of 160 feet above the ground?

Question

The formula ℎ=vt-16⁢^2 gives the height ℎ, in feet, of an object projected into the air with an initial vertical velocity v, in feet per second, after t seconds.

It'll be at 160 feet above the ground at  __ and __ seconds

Answer 1

Using the formula for the height of an object after t seconds,

I believe you have a piece missing from your formula. It is actually

h = vt - 16 t²

v = 112, so

h = 112 t - 16 t²

We are asked what the times are when the height is 160. There are two answers because this height is reached when the object is moving upward, then again when the object is falling.

Put 160 in place of h.

160 = 112t - 16 t²

Move the 160 to the other side and put the terms in descending order of powers.

0 = -16 t² + 112 t - 160

All coefficients are divisible by -16.

0 = t² - 7 t + 10. Factor

0 = (t - 2) (t - 5)

So at t = 2 and at t = 5, the height of the object is 160 feet above the ground.

Answer 2

The actual formula is

h(t) = vt-16t²

For 160 ft

160 = 112t-16t²

Rearranging

16t²-112t+160 = 0

Which is a quadratic equation, so we can use the quadratic equation to solve for t

t = (112 ± √(112²-4(16)(160)))/32 = 2,5

Since this is an inverted parabola, the object will reach 160 feet going up and again coming down.

It'll be at 160 feet above the ground at  2 and 5 seconds.

Answer 3

h(t) = 112t - 16t² = 160; t² - 7t + 10 = 0; t = 2 s or t = 5 s